Converting String to Int with Swift

304

The application basically calculates acceleration by inputting Initial and final velocity and time and then use a formula to calculate acceleration. However, since the values in the text boxes are string, I am unable to convert them to integers.

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel





@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

6

Haven't tried but maybe you could cast the values like var answer1 = Int(txtBox1.text)

– Daniel
Jun 9 '14 at 7:10

If you string is suppose "23.0", then if you cast it to Int("23.0") it will return nil, for this case you first need to cast to Double/Float and then again cast to Int.

– Ariven Nadar
Nov 29 '18 at 14:00

add a comment |

304

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel





@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

6

Haven't tried but maybe you could cast the values like var answer1 = Int(txtBox1.text)

– Daniel
Jun 9 '14 at 7:10

If you string is suppose "23.0", then if you cast it to Int("23.0") it will return nil, for this case you first need to cast to Double/Float and then again cast to Int.

– Ariven Nadar
Nov 29 '18 at 14:00

add a comment |

304

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel





@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel





@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3

ios swift int uitextfield

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

edited Feb 22 '18 at 10:34

ayaio

58.3k20132189

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

asked Jun 9 '14 at 6:57

Marwan Qasem

1,526274

6

Haven't tried but maybe you could cast the values like var answer1 = Int(txtBox1.text)

– Daniel
Jun 9 '14 at 7:10

If you string is suppose "23.0", then if you cast it to Int("23.0") it will return nil, for this case you first need to cast to Double/Float and then again cast to Int.

– Ariven Nadar
Nov 29 '18 at 14:00

add a comment |

6

Haven't tried but maybe you could cast the values like var answer1 = Int(txtBox1.text)

– Daniel
Jun 9 '14 at 7:10

If you string is suppose "23.0", then if you cast it to Int("23.0") it will return nil, for this case you first need to cast to Double/Float and then again cast to Int.

– Ariven Nadar
Nov 29 '18 at 14:00

Haven't tried but maybe you could cast the values like var answer1 = Int(txtBox1.text)

– Daniel
Jun 9 '14 at 7:10

If you string is suppose "23.0", then if you cast it to Int("23.0") it will return nil, for this case you first need to cast to Double/Float and then again cast to Int.

– Ariven Nadar
Nov 29 '18 at 14:00

add a comment |

28 Answers
28

active

oldest

votes

262

Basic Idea, note that this only works in Swift 1.x (check out ParaSara's answer to see how it works in Swift 2.x):

    // toInt returns optional that's why we used a:Int?

    let a:Int? = firstText.text.toInt() // firstText is UITextField

    let b:Int? = secondText.text.toInt() // secondText is UITextField



    // check a and b before unwrapping using !

    if a && b {

        var ans = a! + b!

        answerLabel.text = "Answer is (ans)" // answerLabel ie UILabel

    } else {

        answerLabel.text = "Input values are not numeric"

    }

Update for Swift 4

...

let a:Int? = Int(firstText.text) // firstText is UITextField

let b:Int? = Int(secondText.text) // secondText is UITextField

...

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

Thanks this works. However, I have an issue since I want the numbers to include floats too. Thanks again.

– Marwan Qasem
Jun 12 '14 at 18:15

3

If you need floats (and you really really want Double, not float), toInt() will not do it. Could you use your imagination and the available documentation to find a suitable function?

– gnasher729
Jun 13 '14 at 7:18

4

I get 'NSString' does not have a member named 'toInt'. Any Ideas?

– code ninja
Nov 4 '14 at 11:16

1

NSString and String are two different objects and have different methods. NSString has a method called .intValue

– Byron Coetsee
Jan 22 '15 at 12:24

6

This solution was only right for Swift and not for Swift2. Now you should use: Int(firstText.text)

– gurehbgui
Sep 27 '15 at 10:36

|
show 5 more comments

331

Update Answer for swift 2.0 :

toInt() method is given a error. Because,In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String:

let a:Int? = Int(firstText.text)     // firstText is UITextField  

let b:Int? = Int(secondText.text)   // secondText is UITextField

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

Can i ask why i am getting an error when i omit the '?' char? Why do i need to state 'a' as an optional?

– Manos Serifios
Jun 2 '16 at 20:16

1

@ManosSerifios this discussion may helpful : stackoverflow.com/questions/32621022/…

– Paraneetharan Saravanaperumal
Jun 3 '16 at 5:36

1

@ParaSara Thanks a lot. This make it clear to me.

– Manos Serifios
Jun 3 '16 at 9:55

not truly related but the constructor approach is always preferred and more readable for Int to Strings. "(someInt)" is not good String(someInt) is much easier to read

– Honey
Sep 30 '16 at 16:09

I am printing Int(firstText.text)! and then I still see optional. Why? Have I not unwrapped it?

– Honey
Nov 22 '16 at 17:50

|
show 1 more comment

myString.toInt() - convert the string value into int .

Swift 3.x

If you have an integer hiding inside a string, you can convertby using the integer's constructor, like this:

let myInt = Int(textField.text)

As with other data types (Float and Double) you can also convert by using NSString:

let myString = "556"

let myInt = (myString as NSString).integerValue

edited Jul 5 '17 at 11:41

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

1

this actually answers the question, all of the others tell the OP how to coast an Integer as a String, which is not what he was asking

– aremvee
Jul 8 '15 at 12:19

1

Example for Swift 3?

– Peter Kreinz
Nov 3 '16 at 10:18

pls clarify "latest Swift Versions" for posterity's hoped lack of confusion :)

– Alex Hall
Jul 3 '17 at 21:44

@aremvee do you mean "cast" an integer as a string? And what exactly does this do that answers the question which the other answers don't?

– Alex Hall
Jul 3 '17 at 21:46

add a comment |

Xcode 8.3.2 • Swift 3.1

class IntegerField: UITextField {

    var integer: Int {

        return string.digits.integer

    }

    override func willMove(toSuperview newSuperview: UIView?) {

        addTarget(self, action: #selector(editingChanged), for: .editingChanged)

        keyboardType = .numberPad

        textAlignment = .right

        editingChanged()

    }

    func editingChanged() {

        text = Formatter.decimal.string(for: integer)

        print(integer)

    }

}

Required extensions, structs and initializers:

extension Sequence where Iterator.Element == UnicodeScalar {

    var string: String { return String(String.UnicodeScalarView(self)) }

}



extension Formatter {

    static let decimal = NumberFormatter(numberStyle: .decimal)

}



extension UITextField {

    var string: String { return text ?? "" }

}



extension String {

    private static var digitsPattern = UnicodeScalar("0")..."9"

    var digits: String {

        return unicodeScalars.filter { String.digitsPattern ~= $0 }.string

    }

    var integer: Int { return Int(self) ?? 0 }

}



extension NumberFormatter {

    convenience init(numberStyle: Style) {

        self.init()

        self.numberStyle = numberStyle

    }

}

edited Oct 23 '17 at 17:56

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

add a comment |

You wanna use NSNumberFormatter().numberFromString(yourNumberString). It's great because it returns an an optional that you can then test with an "if let" to determine if the conversion was successful.
eg.

     var myString = "(10)"

     if let myNumber = NSNumberFormatter().numberFromString(myString) {

          var myInt = myNumber.integerValue

          // do what you need to do with myInt

     } else {

          // what ever error code you need to write

     }

edited Jun 9 '15 at 11:38

brainray

7,353852102

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

1

I just changed it to 'myNumber.integerValue' since Xcode 7 won't build with 'intValue'. The latter is of Int32 value

– brainray
Jun 9 '15 at 11:39

add a comment |

swift 4.0

let stringNumber = "123"

let number = Int(stringNumber) //here number is of type "Int?"





//using Forced Unwrapping



if number != nil {         

 //string is converted to Int

}

you could also use Optional Binding other than forced binding.

eg:

  if let number = Int(stringNumber) { 

   // number is of type Int 

  }

answered Dec 12 '17 at 12:32

OOMMEN

17926

add a comment |

//Xcode 8.1 and swift 3.0

We can also handle it by Optional Binding, Simply

let occur = "10"



if let occ = Int(occur) {

        print("By optional binding :", occ*2) // 20



    }

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

add a comment |

Swift 3

The simplest and more secure way is:

@IBOutlet var textFieldA  : UITextField

@IBOutlet var textFieldB  : UITextField

@IBOutlet var answerLabel : UILabel



@IBAction func calculate(sender : AnyObject) {



      if let intValueA = Int(textFieldA),

            let intValueB = Int(textFieldB) {

            let result = intValueA + intValueB

            answerLabel.text = "The acceleration is (result)"

      }

      else {

             answerLabel.text = "The value (intValueA) and/or (intValueB) are not a valid integer value"

      }        

}

Avoid invalid values setting keyboard type to number pad:

 textFieldA.keyboardType = .numberPad

 textFieldB.keyboardType = .numberPad

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

answered Oct 11 '16 at 18:03

torcelly

41147

add a comment |

In Swift 4:

extension String {            

    var numberValue:NSNumber? {

        let formatter = NumberFormatter()

        formatter.numberStyle = .decimal

        return formatter.number(from: self)

    }

}

let someFloat = "12".numberValue

edited Oct 30 '17 at 6:11

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

add a comment |

i have made a simple program, where you have 2 txt field you take input form the user and add them to make it simpler to understand please find the code below.

@IBOutlet weak var result: UILabel!

@IBOutlet weak var one: UITextField!

@IBOutlet weak var two: UITextField!



@IBAction func add(sender: AnyObject) {        

    let count = Int(one.text!)

    let cal = Int(two.text!)

    let sum = count! + cal!

    result.text = "Sum is (sum)"

}

hope this helps.

edited Jan 11 '16 at 18:11

chedabob

5,34821940

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

add a comment |

About int() and Swift 2.x: if you get a nil value after conversion check if you try to convert a string with a big number (for example: 1073741824), in this case try:

let bytesInternet : Int64 = Int64(bytesInternetString)!

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

1

Thank you this worked for my case. Int() was working for me with 16 digit numbers but recently started to fail.

– Ryan Boyd
Apr 30 '16 at 7:03

add a comment |

Latest swift3 this code is simply to convert string to int

let myString = "556"

let myInt = Int(myString)

edited May 14 '17 at 5:40

shim

4,03564679

answered Mar 26 '17 at 17:12

user7642519

add a comment |

Swift 3.0

Try this, you don't need to check for any condition I have done everything just use this function. Send anything string, number, float, double ,etc,. you get a number as a value or 0 if it is unable to convert your value

Function:

func getNumber(number: Any?) -> NSNumber {

    guard let statusNumber:NSNumber = number as? NSNumber else

    {

        guard let statString:String = number as? String else

        {

            return 0

        }

        if let myInteger = Int(statString)

        {

            return NSNumber(value:myInteger)

        }

        else{

            return 0

        }

    }

    return statusNumber

}

Usage:
Add the above function in code and to convert use
let myNumber = getNumber(number: myString)
if the myString has a number or string it returns the number else it returns 0

Example 1:

let number:String = "9834"

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 2:

let number:Double = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 3:

let number = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

edited Nov 7 '17 at 6:34

answered Jan 24 '17 at 6:12

Koushik

7601117

add a comment |

In Swift 4.2 and Xcode 10.1

let string:String = "789"

let intValue:Int = Int(string)!

print(intValue)



let integerValue:Int = 789

let stringValue:String = String(integerValue)

    //OR

//let stringValue:String = "(integerValue)"

print(stringValue)

edited Jan 1 at 5:04

answered Oct 12 '18 at 6:50

iOS

2,76722047

add a comment |

Because a string might contain non-numerical characters you should use a guard to protect the operation. Example:

guard let labelInt:Int = Int(labelString) else {

    return

}



useLabelInt()

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

add a comment |

Use this:

// get the values from text boxes

    let a:Double = firstText.text.bridgeToObjectiveC().doubleValue

    let b:Double = secondText.text.bridgeToObjectiveC().doubleValue



//  we checking against 0.0, because above function return 0.0 if it gets failed to convert

    if (a != 0.0) && (b != 0.0) {

        var ans = a + b

        answerLabel.text = "Answer is (ans)"

    } else {

        answerLabel.text = "Input values are not numberic"

    }

Make your UITextField KeyboardType as DecimalTab from your XIB or storyboard, and remove any if condition for doing any calculation, ie.

var ans = a + b

answerLabel.text = "Answer is (ans)"

Because keyboard type is DecimalPad there is no chance to enter other 0-9 or .

Hope this help !!

edited Jun 13 '14 at 10:03

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

add a comment |

//  To convert user input (i.e string) to int for calculation.I did this , and it works.





    let num:Int? = Int(firstTextField.text!);



    let sum:Int = num!-2



    print(sum);

edited Nov 21 '15 at 10:43

answered Nov 20 '15 at 2:55

Jenny

277

add a comment |

This works for me

var a:Int? = Int(userInput.text!)

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

answered Oct 15 '15 at 21:52

Naishta

6,17934039

How is this different from the solution given in the comment?

– Prune
Oct 15 '15 at 22:25

2

The solution given in the comment is missing "!" at the end , which is expected in Swift 2 and onwards

– Naishta
Oct 22 '15 at 20:45

add a comment |

for Swift3.x

extension String {

    func toInt(defaultValue: Int) -> Int {

        if let n = Int(self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)) {

            return n

        } else {

            return defaultValue

        }

    }

}

answered Apr 28 '17 at 13:02

User9527

2,7852129

add a comment |

In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String

Int(myString)

In your case, you could use Int(textField.text!) insted of textField.text!.toInt()

Swift 1.x

let myString: String = "256"

let myInt: Int? = myString.toInt()

Swift 2.x, 3.x

let myString: String = "256"

let myInt: Int? = Int(myString)

answered Mar 16 '18 at 7:32

Azarmanesh

235

add a comment |

for Alternative solution. You can use extension a native type. You can test with playground.

extension String {

    func add(a: Int) -> Int? {

        if let b = Int(self) {

            return b + a

        }

        else {

            return nil

        }

    }     

}

"2".add(1)

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

add a comment |

My solution is to have a general extension for string to int conversion.

extension String {



 // default: it is a number suitable for your project if the string is not an integer



    func toInt(default: Int) -> Int {

        if let result = Int(self) {

            return result

        }

        else {

            return default  

        }

    }



}

edited Jan 4 '17 at 11:44

answered Jan 4 '17 at 8:28

flame3

1,4581221

add a comment |

@IBAction func calculateAclr(_ sender: Any) {

    if let addition = addition(arrayString: [txtBox1.text, txtBox2.text, txtBox3.text]) {

      print("Answer = (addition)")

      lblAnswer.text = "(addition)"

    }

}



func addition(arrayString: [Any?]) -> Int? {



    var answer:Int?

    for arrayElement in arrayString {

        if let stringValue = arrayElement, let intValue = Int(stringValue)  {

            answer = (answer ?? 0) + intValue

        }

    }



    return answer

}

edited Jul 24 '17 at 9:59

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

add a comment |

As of swift 3, I have to force my #%@! string & int with a "!" otherwise it just doesn't work.

For example:

let prefs = UserDefaults.standard

var counter: String!

counter = prefs.string(forKey:"counter")

print("counter: (counter!)")





var counterInt = Int(counter!)

counterInt = counterInt! + 1

print("counterInt: (counterInt!)")



OUTPUT:

counter: 1

counterInt: 2

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

Can't you just do var counterInt = counter.map { Int($0) } ? Where counter whould be a String?

– Martin
Oct 19 '17 at 16:34

@Martin - No. ? makes its optional and thus adds the word "optional" to the counter string.

– Sam B
Oct 19 '17 at 22:27

IMHO, you should not force unwrap your optionals. Prefer use guard and if let statements

– Martin
Oct 20 '17 at 7:56

add a comment |

Question : string "4.0000" can not be convert into integer using Int("4.000")?

Answer : Int() check string is integer or not if yes then give you integer and otherwise nil. but Float or Double can convert any number string to respective Float or Double without giving nil. Example if you have "45" integer string but using Float("45") gives you 45.0 float value or using Double("4567") gives you 45.0.

Solution : NSString(string: "45.000").integerValue or Int(Float("45.000")!)! to get correct result.

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

add a comment |

I recently got the same issue. Below solution is work for me:

        let strValue = "123"

        let result = (strValue as NSString).integerValue

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

add a comment |

An Int in Swift contains an initializer that accepts a String. It returns an optional Int? as the conversion can fail if the string contains not a number.

By using an if let statement you can validate whether the conversion succeeded.

So your code become something like this:

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel



@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3



    if let intAnswer = Int(txtBox1.text) {

      // Correctly converted

    }

}

answered Mar 1 at 15:50

Patrick

476313

add a comment |

-1

Convert String value to Integer in Swift 4

let strValue:String = "100"

let intValue = strValue as! Int

var intValueFromString:Int = strValue as! Int

or

var intValueFromString = Int(strValue)!

edited Dec 3 '18 at 17:14

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

add a comment |

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28 Answers
28

active

oldest

votes

28 Answers
28

active

oldest

votes

262

Basic Idea, note that this only works in Swift 1.x (check out ParaSara's answer to see how it works in Swift 2.x):

    // toInt returns optional that's why we used a:Int?

    let a:Int? = firstText.text.toInt() // firstText is UITextField

    let b:Int? = secondText.text.toInt() // secondText is UITextField



    // check a and b before unwrapping using !

    if a && b {

        var ans = a! + b!

        answerLabel.text = "Answer is (ans)" // answerLabel ie UILabel

    } else {

        answerLabel.text = "Input values are not numeric"

    }

Update for Swift 4

...

let a:Int? = Int(firstText.text) // firstText is UITextField

let b:Int? = Int(secondText.text) // secondText is UITextField

...

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

Thanks this works. However, I have an issue since I want the numbers to include floats too. Thanks again.

– Marwan Qasem
Jun 12 '14 at 18:15

3

If you need floats (and you really really want Double, not float), toInt() will not do it. Could you use your imagination and the available documentation to find a suitable function?

– gnasher729
Jun 13 '14 at 7:18

4

I get 'NSString' does not have a member named 'toInt'. Any Ideas?

– code ninja
Nov 4 '14 at 11:16

1

NSString and String are two different objects and have different methods. NSString has a method called .intValue

– Byron Coetsee
Jan 22 '15 at 12:24

6

This solution was only right for Swift and not for Swift2. Now you should use: Int(firstText.text)

– gurehbgui
Sep 27 '15 at 10:36

|
show 5 more comments

262

Basic Idea, note that this only works in Swift 1.x (check out ParaSara's answer to see how it works in Swift 2.x):

    // toInt returns optional that's why we used a:Int?

    let a:Int? = firstText.text.toInt() // firstText is UITextField

    let b:Int? = secondText.text.toInt() // secondText is UITextField



    // check a and b before unwrapping using !

    if a && b {

        var ans = a! + b!

        answerLabel.text = "Answer is (ans)" // answerLabel ie UILabel

    } else {

        answerLabel.text = "Input values are not numeric"

    }

Update for Swift 4

...

let a:Int? = Int(firstText.text) // firstText is UITextField

let b:Int? = Int(secondText.text) // secondText is UITextField

...

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

Thanks this works. However, I have an issue since I want the numbers to include floats too. Thanks again.

– Marwan Qasem
Jun 12 '14 at 18:15

3

If you need floats (and you really really want Double, not float), toInt() will not do it. Could you use your imagination and the available documentation to find a suitable function?

– gnasher729
Jun 13 '14 at 7:18

4

I get 'NSString' does not have a member named 'toInt'. Any Ideas?

– code ninja
Nov 4 '14 at 11:16

1

NSString and String are two different objects and have different methods. NSString has a method called .intValue

– Byron Coetsee
Jan 22 '15 at 12:24

6

This solution was only right for Swift and not for Swift2. Now you should use: Int(firstText.text)

– gurehbgui
Sep 27 '15 at 10:36

|
show 5 more comments

262

Basic Idea, note that this only works in Swift 1.x (check out ParaSara's answer to see how it works in Swift 2.x):

    // toInt returns optional that's why we used a:Int?

    let a:Int? = firstText.text.toInt() // firstText is UITextField

    let b:Int? = secondText.text.toInt() // secondText is UITextField



    // check a and b before unwrapping using !

    if a && b {

        var ans = a! + b!

        answerLabel.text = "Answer is (ans)" // answerLabel ie UILabel

    } else {

        answerLabel.text = "Input values are not numeric"

    }

Update for Swift 4

...

let a:Int? = Int(firstText.text) // firstText is UITextField

let b:Int? = Int(secondText.text) // secondText is UITextField

...

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

Basic Idea, note that this only works in Swift 1.x (check out ParaSara's answer to see how it works in Swift 2.x):

    // toInt returns optional that's why we used a:Int?

    let a:Int? = firstText.text.toInt() // firstText is UITextField

    let b:Int? = secondText.text.toInt() // secondText is UITextField



    // check a and b before unwrapping using !

    if a && b {

        var ans = a! + b!

        answerLabel.text = "Answer is (ans)" // answerLabel ie UILabel

    } else {

        answerLabel.text = "Input values are not numeric"

    }

Update for Swift 4

...

let a:Int? = Int(firstText.text) // firstText is UITextField

let b:Int? = Int(secondText.text) // secondText is UITextField

...

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

edited Mar 7 '18 at 10:05

Anton Belousov

9991126

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

answered Jun 12 '14 at 11:59

Narendar Singh Saini

2,4101814

Thanks this works. However, I have an issue since I want the numbers to include floats too. Thanks again.

– Marwan Qasem
Jun 12 '14 at 18:15

3

If you need floats (and you really really want Double, not float), toInt() will not do it. Could you use your imagination and the available documentation to find a suitable function?

– gnasher729
Jun 13 '14 at 7:18

4

I get 'NSString' does not have a member named 'toInt'. Any Ideas?

– code ninja
Nov 4 '14 at 11:16

1

NSString and String are two different objects and have different methods. NSString has a method called .intValue

– Byron Coetsee
Jan 22 '15 at 12:24

6

This solution was only right for Swift and not for Swift2. Now you should use: Int(firstText.text)

– gurehbgui
Sep 27 '15 at 10:36

|
show 5 more comments

Thanks this works. However, I have an issue since I want the numbers to include floats too. Thanks again.

– Marwan Qasem
Jun 12 '14 at 18:15

3

If you need floats (and you really really want Double, not float), toInt() will not do it. Could you use your imagination and the available documentation to find a suitable function?

– gnasher729
Jun 13 '14 at 7:18

4

I get 'NSString' does not have a member named 'toInt'. Any Ideas?

– code ninja
Nov 4 '14 at 11:16

1

NSString and String are two different objects and have different methods. NSString has a method called .intValue

– Byron Coetsee
Jan 22 '15 at 12:24

6

This solution was only right for Swift and not for Swift2. Now you should use: Int(firstText.text)

– gurehbgui
Sep 27 '15 at 10:36

Thanks this works. However, I have an issue since I want the numbers to include floats too. Thanks again.

– Marwan Qasem
Jun 12 '14 at 18:15

If you need floats (and you really really want Double, not float), toInt() will not do it. Could you use your imagination and the available documentation to find a suitable function?

– gnasher729
Jun 13 '14 at 7:18

I get 'NSString' does not have a member named 'toInt'. Any Ideas?

– code ninja
Nov 4 '14 at 11:16

NSString and String are two different objects and have different methods. NSString has a method called .intValue

– Byron Coetsee
Jan 22 '15 at 12:24

This solution was only right for Swift and not for Swift2. Now you should use: Int(firstText.text)

– gurehbgui
Sep 27 '15 at 10:36

|
show 5 more comments

331

Update Answer for swift 2.0 :

toInt() method is given a error. Because,In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String:

let a:Int? = Int(firstText.text)     // firstText is UITextField  

let b:Int? = Int(secondText.text)   // secondText is UITextField

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

Can i ask why i am getting an error when i omit the '?' char? Why do i need to state 'a' as an optional?

– Manos Serifios
Jun 2 '16 at 20:16

1

@ManosSerifios this discussion may helpful : stackoverflow.com/questions/32621022/…

– Paraneetharan Saravanaperumal
Jun 3 '16 at 5:36

1

@ParaSara Thanks a lot. This make it clear to me.

– Manos Serifios
Jun 3 '16 at 9:55

not truly related but the constructor approach is always preferred and more readable for Int to Strings. "(someInt)" is not good String(someInt) is much easier to read

– Honey
Sep 30 '16 at 16:09

I am printing Int(firstText.text)! and then I still see optional. Why? Have I not unwrapped it?

– Honey
Nov 22 '16 at 17:50

|
show 1 more comment

331

Update Answer for swift 2.0 :

toInt() method is given a error. Because,In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String:

let a:Int? = Int(firstText.text)     // firstText is UITextField  

let b:Int? = Int(secondText.text)   // secondText is UITextField

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

Can i ask why i am getting an error when i omit the '?' char? Why do i need to state 'a' as an optional?

– Manos Serifios
Jun 2 '16 at 20:16

1

@ManosSerifios this discussion may helpful : stackoverflow.com/questions/32621022/…

– Paraneetharan Saravanaperumal
Jun 3 '16 at 5:36

1

@ParaSara Thanks a lot. This make it clear to me.

– Manos Serifios
Jun 3 '16 at 9:55

not truly related but the constructor approach is always preferred and more readable for Int to Strings. "(someInt)" is not good String(someInt) is much easier to read

– Honey
Sep 30 '16 at 16:09

I am printing Int(firstText.text)! and then I still see optional. Why? Have I not unwrapped it?

– Honey
Nov 22 '16 at 17:50

|
show 1 more comment

331

Update Answer for swift 2.0 :

toInt() method is given a error. Because,In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String:

let a:Int? = Int(firstText.text)     // firstText is UITextField  

let b:Int? = Int(secondText.text)   // secondText is UITextField

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

Update Answer for swift 2.0 :

toInt() method is given a error. Because,In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String:

let a:Int? = Int(firstText.text)     // firstText is UITextField  

let b:Int? = Int(secondText.text)   // secondText is UITextField

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

edited Dec 2 '16 at 5:16

rptwsthi

8,56885393

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

answered Jun 11 '15 at 6:22

Paraneetharan Saravanaperumal

4,13011024

Can i ask why i am getting an error when i omit the '?' char? Why do i need to state 'a' as an optional?

– Manos Serifios
Jun 2 '16 at 20:16

1

@ManosSerifios this discussion may helpful : stackoverflow.com/questions/32621022/…

– Paraneetharan Saravanaperumal
Jun 3 '16 at 5:36

1

@ParaSara Thanks a lot. This make it clear to me.

– Manos Serifios
Jun 3 '16 at 9:55

not truly related but the constructor approach is always preferred and more readable for Int to Strings. "(someInt)" is not good String(someInt) is much easier to read

– Honey
Sep 30 '16 at 16:09

I am printing Int(firstText.text)! and then I still see optional. Why? Have I not unwrapped it?

– Honey
Nov 22 '16 at 17:50

|
show 1 more comment

Can i ask why i am getting an error when i omit the '?' char? Why do i need to state 'a' as an optional?

– Manos Serifios
Jun 2 '16 at 20:16

1

@ManosSerifios this discussion may helpful : stackoverflow.com/questions/32621022/…

– Paraneetharan Saravanaperumal
Jun 3 '16 at 5:36

1

@ParaSara Thanks a lot. This make it clear to me.

– Manos Serifios
Jun 3 '16 at 9:55

not truly related but the constructor approach is always preferred and more readable for Int to Strings. "(someInt)" is not good String(someInt) is much easier to read

– Honey
Sep 30 '16 at 16:09

I am printing Int(firstText.text)! and then I still see optional. Why? Have I not unwrapped it?

– Honey
Nov 22 '16 at 17:50

Can i ask why i am getting an error when i omit the '?' char? Why do i need to state 'a' as an optional?

– Manos Serifios
Jun 2 '16 at 20:16

@ManosSerifios this discussion may helpful : stackoverflow.com/questions/32621022/…

– Paraneetharan Saravanaperumal
Jun 3 '16 at 5:36

@ParaSara Thanks a lot. This make it clear to me.

– Manos Serifios
Jun 3 '16 at 9:55

not truly related but the constructor approach is always preferred and more readable for Int to Strings. "(someInt)" is not good String(someInt) is much easier to read

– Honey
Sep 30 '16 at 16:09

I am printing Int(firstText.text)! and then I still see optional. Why? Have I not unwrapped it?

– Honey
Nov 22 '16 at 17:50

|
show 1 more comment

myString.toInt() - convert the string value into int .

Swift 3.x

If you have an integer hiding inside a string, you can convertby using the integer's constructor, like this:

let myInt = Int(textField.text)

As with other data types (Float and Double) you can also convert by using NSString:

let myString = "556"

let myInt = (myString as NSString).integerValue

edited Jul 5 '17 at 11:41

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

1

this actually answers the question, all of the others tell the OP how to coast an Integer as a String, which is not what he was asking

– aremvee
Jul 8 '15 at 12:19

1

Example for Swift 3?

– Peter Kreinz
Nov 3 '16 at 10:18

pls clarify "latest Swift Versions" for posterity's hoped lack of confusion :)

– Alex Hall
Jul 3 '17 at 21:44

@aremvee do you mean "cast" an integer as a string? And what exactly does this do that answers the question which the other answers don't?

– Alex Hall
Jul 3 '17 at 21:46

add a comment |

myString.toInt() - convert the string value into int .

Swift 3.x

If you have an integer hiding inside a string, you can convertby using the integer's constructor, like this:

let myInt = Int(textField.text)

As with other data types (Float and Double) you can also convert by using NSString:

let myString = "556"

let myInt = (myString as NSString).integerValue

edited Jul 5 '17 at 11:41

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

1

this actually answers the question, all of the others tell the OP how to coast an Integer as a String, which is not what he was asking

– aremvee
Jul 8 '15 at 12:19

1

Example for Swift 3?

– Peter Kreinz
Nov 3 '16 at 10:18

pls clarify "latest Swift Versions" for posterity's hoped lack of confusion :)

– Alex Hall
Jul 3 '17 at 21:44

@aremvee do you mean "cast" an integer as a string? And what exactly does this do that answers the question which the other answers don't?

– Alex Hall
Jul 3 '17 at 21:46

add a comment |

myString.toInt() - convert the string value into int .

Swift 3.x

If you have an integer hiding inside a string, you can convertby using the integer's constructor, like this:

let myInt = Int(textField.text)

As with other data types (Float and Double) you can also convert by using NSString:

let myString = "556"

let myInt = (myString as NSString).integerValue

edited Jul 5 '17 at 11:41

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

myString.toInt() - convert the string value into int .

Swift 3.x

If you have an integer hiding inside a string, you can convertby using the integer's constructor, like this:

let myInt = Int(textField.text)

As with other data types (Float and Double) you can also convert by using NSString:

let myString = "556"

let myInt = (myString as NSString).integerValue

edited Jul 5 '17 at 11:41

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

edited Jul 5 '17 at 11:41

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

answered Jun 9 '14 at 7:13

Kumar KL

11.4k93354

1

this actually answers the question, all of the others tell the OP how to coast an Integer as a String, which is not what he was asking

– aremvee
Jul 8 '15 at 12:19

1

Example for Swift 3?

– Peter Kreinz
Nov 3 '16 at 10:18

pls clarify "latest Swift Versions" for posterity's hoped lack of confusion :)

– Alex Hall
Jul 3 '17 at 21:44

@aremvee do you mean "cast" an integer as a string? And what exactly does this do that answers the question which the other answers don't?

– Alex Hall
Jul 3 '17 at 21:46

add a comment |

1

this actually answers the question, all of the others tell the OP how to coast an Integer as a String, which is not what he was asking

– aremvee
Jul 8 '15 at 12:19

1

Example for Swift 3?

– Peter Kreinz
Nov 3 '16 at 10:18

pls clarify "latest Swift Versions" for posterity's hoped lack of confusion :)

– Alex Hall
Jul 3 '17 at 21:44

@aremvee do you mean "cast" an integer as a string? And what exactly does this do that answers the question which the other answers don't?

– Alex Hall
Jul 3 '17 at 21:46

this actually answers the question, all of the others tell the OP how to coast an Integer as a String, which is not what he was asking

– aremvee
Jul 8 '15 at 12:19

Example for Swift 3?

– Peter Kreinz
Nov 3 '16 at 10:18

pls clarify "latest Swift Versions" for posterity's hoped lack of confusion :)

– Alex Hall
Jul 3 '17 at 21:44

@aremvee do you mean "cast" an integer as a string? And what exactly does this do that answers the question which the other answers don't?

– Alex Hall
Jul 3 '17 at 21:46

add a comment |

Xcode 8.3.2 • Swift 3.1

class IntegerField: UITextField {

    var integer: Int {

        return string.digits.integer

    }

    override func willMove(toSuperview newSuperview: UIView?) {

        addTarget(self, action: #selector(editingChanged), for: .editingChanged)

        keyboardType = .numberPad

        textAlignment = .right

        editingChanged()

    }

    func editingChanged() {

        text = Formatter.decimal.string(for: integer)

        print(integer)

    }

}

Required extensions, structs and initializers:

extension Sequence where Iterator.Element == UnicodeScalar {

    var string: String { return String(String.UnicodeScalarView(self)) }

}



extension Formatter {

    static let decimal = NumberFormatter(numberStyle: .decimal)

}



extension UITextField {

    var string: String { return text ?? "" }

}



extension String {

    private static var digitsPattern = UnicodeScalar("0")..."9"

    var digits: String {

        return unicodeScalars.filter { String.digitsPattern ~= $0 }.string

    }

    var integer: Int { return Int(self) ?? 0 }

}



extension NumberFormatter {

    convenience init(numberStyle: Style) {

        self.init()

        self.numberStyle = numberStyle

    }

}

edited Oct 23 '17 at 17:56

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

add a comment |

Xcode 8.3.2 • Swift 3.1

class IntegerField: UITextField {

    var integer: Int {

        return string.digits.integer

    }

    override func willMove(toSuperview newSuperview: UIView?) {

        addTarget(self, action: #selector(editingChanged), for: .editingChanged)

        keyboardType = .numberPad

        textAlignment = .right

        editingChanged()

    }

    func editingChanged() {

        text = Formatter.decimal.string(for: integer)

        print(integer)

    }

}

Required extensions, structs and initializers:

extension Sequence where Iterator.Element == UnicodeScalar {

    var string: String { return String(String.UnicodeScalarView(self)) }

}



extension Formatter {

    static let decimal = NumberFormatter(numberStyle: .decimal)

}



extension UITextField {

    var string: String { return text ?? "" }

}



extension String {

    private static var digitsPattern = UnicodeScalar("0")..."9"

    var digits: String {

        return unicodeScalars.filter { String.digitsPattern ~= $0 }.string

    }

    var integer: Int { return Int(self) ?? 0 }

}



extension NumberFormatter {

    convenience init(numberStyle: Style) {

        self.init()

        self.numberStyle = numberStyle

    }

}

edited Oct 23 '17 at 17:56

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

add a comment |

Xcode 8.3.2 • Swift 3.1

class IntegerField: UITextField {

    var integer: Int {

        return string.digits.integer

    }

    override func willMove(toSuperview newSuperview: UIView?) {

        addTarget(self, action: #selector(editingChanged), for: .editingChanged)

        keyboardType = .numberPad

        textAlignment = .right

        editingChanged()

    }

    func editingChanged() {

        text = Formatter.decimal.string(for: integer)

        print(integer)

    }

}

Required extensions, structs and initializers:

extension Sequence where Iterator.Element == UnicodeScalar {

    var string: String { return String(String.UnicodeScalarView(self)) }

}



extension Formatter {

    static let decimal = NumberFormatter(numberStyle: .decimal)

}



extension UITextField {

    var string: String { return text ?? "" }

}



extension String {

    private static var digitsPattern = UnicodeScalar("0")..."9"

    var digits: String {

        return unicodeScalars.filter { String.digitsPattern ~= $0 }.string

    }

    var integer: Int { return Int(self) ?? 0 }

}



extension NumberFormatter {

    convenience init(numberStyle: Style) {

        self.init()

        self.numberStyle = numberStyle

    }

}

edited Oct 23 '17 at 17:56

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

Xcode 8.3.2 • Swift 3.1

class IntegerField: UITextField {

    var integer: Int {

        return string.digits.integer

    }

    override func willMove(toSuperview newSuperview: UIView?) {

        addTarget(self, action: #selector(editingChanged), for: .editingChanged)

        keyboardType = .numberPad

        textAlignment = .right

        editingChanged()

    }

    func editingChanged() {

        text = Formatter.decimal.string(for: integer)

        print(integer)

    }

}

Required extensions, structs and initializers:

extension Sequence where Iterator.Element == UnicodeScalar {

    var string: String { return String(String.UnicodeScalarView(self)) }

}



extension Formatter {

    static let decimal = NumberFormatter(numberStyle: .decimal)

}



extension UITextField {

    var string: String { return text ?? "" }

}



extension String {

    private static var digitsPattern = UnicodeScalar("0")..."9"

    var digits: String {

        return unicodeScalars.filter { String.digitsPattern ~= $0 }.string

    }

    var integer: Int { return Int(self) ?? 0 }

}



extension NumberFormatter {

    convenience init(numberStyle: Style) {

        self.init()

        self.numberStyle = numberStyle

    }

}

edited Oct 23 '17 at 17:56

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

edited Oct 23 '17 at 17:56

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

answered Dec 15 '15 at 16:39

Leo Dabus

135k32277351

add a comment |

     var myString = "(10)"

     if let myNumber = NSNumberFormatter().numberFromString(myString) {

          var myInt = myNumber.integerValue

          // do what you need to do with myInt

     } else {

          // what ever error code you need to write

     }

edited Jun 9 '15 at 11:38

brainray

7,353852102

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

1

I just changed it to 'myNumber.integerValue' since Xcode 7 won't build with 'intValue'. The latter is of Int32 value

– brainray
Jun 9 '15 at 11:39

add a comment |

     var myString = "(10)"

     if let myNumber = NSNumberFormatter().numberFromString(myString) {

          var myInt = myNumber.integerValue

          // do what you need to do with myInt

     } else {

          // what ever error code you need to write

     }

edited Jun 9 '15 at 11:38

brainray

7,353852102

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

1

I just changed it to 'myNumber.integerValue' since Xcode 7 won't build with 'intValue'. The latter is of Int32 value

– brainray
Jun 9 '15 at 11:39

add a comment |

     var myString = "(10)"

     if let myNumber = NSNumberFormatter().numberFromString(myString) {

          var myInt = myNumber.integerValue

          // do what you need to do with myInt

     } else {

          // what ever error code you need to write

     }

edited Jun 9 '15 at 11:38

brainray

7,353852102

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

     var myString = "(10)"

     if let myNumber = NSNumberFormatter().numberFromString(myString) {

          var myInt = myNumber.integerValue

          // do what you need to do with myInt

     } else {

          // what ever error code you need to write

     }

edited Jun 9 '15 at 11:38

brainray

7,353852102

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

edited Jun 9 '15 at 11:38

brainray

7,353852102

edited Jun 9 '15 at 11:38

brainray

7,353852102

edited Jun 9 '15 at 11:38

brainray

7,353852102

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

answered Feb 27 '15 at 14:03

Abaho Katabarwa

22922

1

I just changed it to 'myNumber.integerValue' since Xcode 7 won't build with 'intValue'. The latter is of Int32 value

– brainray
Jun 9 '15 at 11:39

add a comment |

1

I just changed it to 'myNumber.integerValue' since Xcode 7 won't build with 'intValue'. The latter is of Int32 value

– brainray
Jun 9 '15 at 11:39

I just changed it to 'myNumber.integerValue' since Xcode 7 won't build with 'intValue'. The latter is of Int32 value

– brainray
Jun 9 '15 at 11:39

add a comment |

swift 4.0

let stringNumber = "123"

let number = Int(stringNumber) //here number is of type "Int?"





//using Forced Unwrapping



if number != nil {         

 //string is converted to Int

}

you could also use Optional Binding other than forced binding.

eg:

  if let number = Int(stringNumber) { 

   // number is of type Int 

  }

answered Dec 12 '17 at 12:32

OOMMEN

17926

add a comment |

swift 4.0

let stringNumber = "123"

let number = Int(stringNumber) //here number is of type "Int?"





//using Forced Unwrapping



if number != nil {         

 //string is converted to Int

}

you could also use Optional Binding other than forced binding.

eg:

  if let number = Int(stringNumber) { 

   // number is of type Int 

  }

answered Dec 12 '17 at 12:32

OOMMEN

17926

add a comment |

swift 4.0

let stringNumber = "123"

let number = Int(stringNumber) //here number is of type "Int?"





//using Forced Unwrapping



if number != nil {         

 //string is converted to Int

}

you could also use Optional Binding other than forced binding.

eg:

  if let number = Int(stringNumber) { 

   // number is of type Int 

  }

answered Dec 12 '17 at 12:32

OOMMEN

17926

swift 4.0

let stringNumber = "123"

let number = Int(stringNumber) //here number is of type "Int?"





//using Forced Unwrapping



if number != nil {         

 //string is converted to Int

}

you could also use Optional Binding other than forced binding.

eg:

  if let number = Int(stringNumber) { 

   // number is of type Int 

  }

answered Dec 12 '17 at 12:32

OOMMEN

17926

answered Dec 12 '17 at 12:32

OOMMEN

17926

answered Dec 12 '17 at 12:32

OOMMEN

17926

answered Dec 12 '17 at 12:32

OOMMEN

17926

add a comment |

//Xcode 8.1 and swift 3.0

We can also handle it by Optional Binding, Simply

let occur = "10"



if let occ = Int(occur) {

        print("By optional binding :", occ*2) // 20



    }

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

add a comment |

//Xcode 8.1 and swift 3.0

We can also handle it by Optional Binding, Simply

let occur = "10"



if let occ = Int(occur) {

        print("By optional binding :", occ*2) // 20



    }

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

add a comment |

//Xcode 8.1 and swift 3.0

We can also handle it by Optional Binding, Simply

let occur = "10"



if let occ = Int(occur) {

        print("By optional binding :", occ*2) // 20



    }

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

//Xcode 8.1 and swift 3.0

We can also handle it by Optional Binding, Simply

let occur = "10"



if let occ = Int(occur) {

        print("By optional binding :", occ*2) // 20



    }

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

edited May 25 '17 at 14:33

CupawnTae

11.3k22251

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

answered Nov 25 '16 at 7:47

Pankaj Nigam

17816

add a comment |

Swift 3

The simplest and more secure way is:

@IBOutlet var textFieldA  : UITextField

@IBOutlet var textFieldB  : UITextField

@IBOutlet var answerLabel : UILabel



@IBAction func calculate(sender : AnyObject) {



      if let intValueA = Int(textFieldA),

            let intValueB = Int(textFieldB) {

            let result = intValueA + intValueB

            answerLabel.text = "The acceleration is (result)"

      }

      else {

             answerLabel.text = "The value (intValueA) and/or (intValueB) are not a valid integer value"

      }        

}

Avoid invalid values setting keyboard type to number pad:

 textFieldA.keyboardType = .numberPad

 textFieldB.keyboardType = .numberPad

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

answered Oct 11 '16 at 18:03

torcelly

41147

add a comment |

Swift 3

The simplest and more secure way is:

@IBOutlet var textFieldA  : UITextField

@IBOutlet var textFieldB  : UITextField

@IBOutlet var answerLabel : UILabel



@IBAction func calculate(sender : AnyObject) {



      if let intValueA = Int(textFieldA),

            let intValueB = Int(textFieldB) {

            let result = intValueA + intValueB

            answerLabel.text = "The acceleration is (result)"

      }

      else {

             answerLabel.text = "The value (intValueA) and/or (intValueB) are not a valid integer value"

      }        

}

Avoid invalid values setting keyboard type to number pad:

 textFieldA.keyboardType = .numberPad

 textFieldB.keyboardType = .numberPad

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

answered Oct 11 '16 at 18:03

torcelly

41147

add a comment |

Swift 3

The simplest and more secure way is:

@IBOutlet var textFieldA  : UITextField

@IBOutlet var textFieldB  : UITextField

@IBOutlet var answerLabel : UILabel



@IBAction func calculate(sender : AnyObject) {



      if let intValueA = Int(textFieldA),

            let intValueB = Int(textFieldB) {

            let result = intValueA + intValueB

            answerLabel.text = "The acceleration is (result)"

      }

      else {

             answerLabel.text = "The value (intValueA) and/or (intValueB) are not a valid integer value"

      }        

}

Avoid invalid values setting keyboard type to number pad:

 textFieldA.keyboardType = .numberPad

 textFieldB.keyboardType = .numberPad

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

answered Oct 11 '16 at 18:03

torcelly

41147

Swift 3

The simplest and more secure way is:

@IBOutlet var textFieldA  : UITextField

@IBOutlet var textFieldB  : UITextField

@IBOutlet var answerLabel : UILabel



@IBAction func calculate(sender : AnyObject) {



      if let intValueA = Int(textFieldA),

            let intValueB = Int(textFieldB) {

            let result = intValueA + intValueB

            answerLabel.text = "The acceleration is (result)"

      }

      else {

             answerLabel.text = "The value (intValueA) and/or (intValueB) are not a valid integer value"

      }        

}

Avoid invalid values setting keyboard type to number pad:

 textFieldA.keyboardType = .numberPad

 textFieldB.keyboardType = .numberPad

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

answered Oct 11 '16 at 18:03

torcelly

41147

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

edited Mar 14 '17 at 4:37

John R Perry

1,7781637

answered Oct 11 '16 at 18:03

torcelly

41147

answered Oct 11 '16 at 18:03

torcelly

41147

answered Oct 11 '16 at 18:03

torcelly

41147

add a comment |

In Swift 4:

extension String {            

    var numberValue:NSNumber? {

        let formatter = NumberFormatter()

        formatter.numberStyle = .decimal

        return formatter.number(from: self)

    }

}

let someFloat = "12".numberValue

edited Oct 30 '17 at 6:11

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

add a comment |

In Swift 4:

extension String {            

    var numberValue:NSNumber? {

        let formatter = NumberFormatter()

        formatter.numberStyle = .decimal

        return formatter.number(from: self)

    }

}

let someFloat = "12".numberValue

edited Oct 30 '17 at 6:11

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

add a comment |

In Swift 4:

extension String {            

    var numberValue:NSNumber? {

        let formatter = NumberFormatter()

        formatter.numberStyle = .decimal

        return formatter.number(from: self)

    }

}

let someFloat = "12".numberValue

edited Oct 30 '17 at 6:11

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

In Swift 4:

extension String {            

    var numberValue:NSNumber? {

        let formatter = NumberFormatter()

        formatter.numberStyle = .decimal

        return formatter.number(from: self)

    }

}

let someFloat = "12".numberValue

edited Oct 30 '17 at 6:11

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

edited Oct 30 '17 at 6:11

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

answered Oct 12 '17 at 18:53

Ankit garg

1,2111214

add a comment |

i have made a simple program, where you have 2 txt field you take input form the user and add them to make it simpler to understand please find the code below.

@IBOutlet weak var result: UILabel!

@IBOutlet weak var one: UITextField!

@IBOutlet weak var two: UITextField!



@IBAction func add(sender: AnyObject) {        

    let count = Int(one.text!)

    let cal = Int(two.text!)

    let sum = count! + cal!

    result.text = "Sum is (sum)"

}

hope this helps.

edited Jan 11 '16 at 18:11

chedabob

5,34821940

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

add a comment |

i have made a simple program, where you have 2 txt field you take input form the user and add them to make it simpler to understand please find the code below.

@IBOutlet weak var result: UILabel!

@IBOutlet weak var one: UITextField!

@IBOutlet weak var two: UITextField!



@IBAction func add(sender: AnyObject) {        

    let count = Int(one.text!)

    let cal = Int(two.text!)

    let sum = count! + cal!

    result.text = "Sum is (sum)"

}

hope this helps.

edited Jan 11 '16 at 18:11

chedabob

5,34821940

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

add a comment |

i have made a simple program, where you have 2 txt field you take input form the user and add them to make it simpler to understand please find the code below.

@IBOutlet weak var result: UILabel!

@IBOutlet weak var one: UITextField!

@IBOutlet weak var two: UITextField!



@IBAction func add(sender: AnyObject) {        

    let count = Int(one.text!)

    let cal = Int(two.text!)

    let sum = count! + cal!

    result.text = "Sum is (sum)"

}

hope this helps.

edited Jan 11 '16 at 18:11

chedabob

5,34821940

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

i have made a simple program, where you have 2 txt field you take input form the user and add them to make it simpler to understand please find the code below.

@IBOutlet weak var result: UILabel!

@IBOutlet weak var one: UITextField!

@IBOutlet weak var two: UITextField!



@IBAction func add(sender: AnyObject) {        

    let count = Int(one.text!)

    let cal = Int(two.text!)

    let sum = count! + cal!

    result.text = "Sum is (sum)"

}

hope this helps.

edited Jan 11 '16 at 18:11

chedabob

5,34821940

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

edited Jan 11 '16 at 18:11

chedabob

5,34821940

edited Jan 11 '16 at 18:11

chedabob

5,34821940

edited Jan 11 '16 at 18:11

chedabob

5,34821940

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

answered Jan 11 '16 at 18:07

AliRaza Qureshi

411

add a comment |

About int() and Swift 2.x: if you get a nil value after conversion check if you try to convert a string with a big number (for example: 1073741824), in this case try:

let bytesInternet : Int64 = Int64(bytesInternetString)!

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

1

Thank you this worked for my case. Int() was working for me with 16 digit numbers but recently started to fail.

– Ryan Boyd
Apr 30 '16 at 7:03

add a comment |

About int() and Swift 2.x: if you get a nil value after conversion check if you try to convert a string with a big number (for example: 1073741824), in this case try:

let bytesInternet : Int64 = Int64(bytesInternetString)!

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

1

Thank you this worked for my case. Int() was working for me with 16 digit numbers but recently started to fail.

– Ryan Boyd
Apr 30 '16 at 7:03

add a comment |

About int() and Swift 2.x: if you get a nil value after conversion check if you try to convert a string with a big number (for example: 1073741824), in this case try:

let bytesInternet : Int64 = Int64(bytesInternetString)!

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

About int() and Swift 2.x: if you get a nil value after conversion check if you try to convert a string with a big number (for example: 1073741824), in this case try:

let bytesInternet : Int64 = Int64(bytesInternetString)!

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

answered Oct 30 '15 at 18:34

Alessandro Ornano

23.1k106585

1

Thank you this worked for my case. Int() was working for me with 16 digit numbers but recently started to fail.

– Ryan Boyd
Apr 30 '16 at 7:03

add a comment |

1

Thank you this worked for my case. Int() was working for me with 16 digit numbers but recently started to fail.

– Ryan Boyd
Apr 30 '16 at 7:03

Thank you this worked for my case. Int() was working for me with 16 digit numbers but recently started to fail.

– Ryan Boyd
Apr 30 '16 at 7:03

add a comment |

Latest swift3 this code is simply to convert string to int

let myString = "556"

let myInt = Int(myString)

edited May 14 '17 at 5:40

shim

4,03564679

answered Mar 26 '17 at 17:12

user7642519

add a comment |

Latest swift3 this code is simply to convert string to int

let myString = "556"

let myInt = Int(myString)

edited May 14 '17 at 5:40

shim

4,03564679

answered Mar 26 '17 at 17:12

user7642519

add a comment |

Latest swift3 this code is simply to convert string to int

let myString = "556"

let myInt = Int(myString)

edited May 14 '17 at 5:40

shim

4,03564679

answered Mar 26 '17 at 17:12

user7642519

Latest swift3 this code is simply to convert string to int

let myString = "556"

let myInt = Int(myString)

edited May 14 '17 at 5:40

shim

4,03564679

answered Mar 26 '17 at 17:12

user7642519

edited May 14 '17 at 5:40

shim

4,03564679

edited May 14 '17 at 5:40

shim

4,03564679

edited May 14 '17 at 5:40

shim

4,03564679

answered Mar 26 '17 at 17:12

user7642519

answered Mar 26 '17 at 17:12

user7642519

answered Mar 26 '17 at 17:12

user7642519

add a comment |

Swift 3.0

Function:

func getNumber(number: Any?) -> NSNumber {

    guard let statusNumber:NSNumber = number as? NSNumber else

    {

        guard let statString:String = number as? String else

        {

            return 0

        }

        if let myInteger = Int(statString)

        {

            return NSNumber(value:myInteger)

        }

        else{

            return 0

        }

    }

    return statusNumber

}

Usage:
Add the above function in code and to convert use
let myNumber = getNumber(number: myString)
if the myString has a number or string it returns the number else it returns 0

Example 1:

let number:String = "9834"

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 2:

let number:Double = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 3:

let number = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

edited Nov 7 '17 at 6:34

answered Jan 24 '17 at 6:12

Koushik

7601117

add a comment |

Swift 3.0

Function:

func getNumber(number: Any?) -> NSNumber {

    guard let statusNumber:NSNumber = number as? NSNumber else

    {

        guard let statString:String = number as? String else

        {

            return 0

        }

        if let myInteger = Int(statString)

        {

            return NSNumber(value:myInteger)

        }

        else{

            return 0

        }

    }

    return statusNumber

}

Usage:
Add the above function in code and to convert use
let myNumber = getNumber(number: myString)
if the myString has a number or string it returns the number else it returns 0

Example 1:

let number:String = "9834"

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 2:

let number:Double = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 3:

let number = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

edited Nov 7 '17 at 6:34

answered Jan 24 '17 at 6:12

Koushik

7601117

add a comment |

Swift 3.0

Function:

func getNumber(number: Any?) -> NSNumber {

    guard let statusNumber:NSNumber = number as? NSNumber else

    {

        guard let statString:String = number as? String else

        {

            return 0

        }

        if let myInteger = Int(statString)

        {

            return NSNumber(value:myInteger)

        }

        else{

            return 0

        }

    }

    return statusNumber

}

Usage:
Add the above function in code and to convert use
let myNumber = getNumber(number: myString)
if the myString has a number or string it returns the number else it returns 0

Example 1:

let number:String = "9834"

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 2:

let number:Double = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 3:

let number = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

edited Nov 7 '17 at 6:34

answered Jan 24 '17 at 6:12

Koushik

7601117

Swift 3.0

Function:

func getNumber(number: Any?) -> NSNumber {

    guard let statusNumber:NSNumber = number as? NSNumber else

    {

        guard let statString:String = number as? String else

        {

            return 0

        }

        if let myInteger = Int(statString)

        {

            return NSNumber(value:myInteger)

        }

        else{

            return 0

        }

    }

    return statusNumber

}

Usage:
Add the above function in code and to convert use
let myNumber = getNumber(number: myString)
if the myString has a number or string it returns the number else it returns 0

Example 1:

let number:String = "9834"

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 2:

let number:Double = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

Example 3:

let number = 9834

print("printing number (getNumber(number: number))")

Output: printing number 9834

edited Nov 7 '17 at 6:34

answered Jan 24 '17 at 6:12

Koushik

7601117

edited Nov 7 '17 at 6:34

answered Jan 24 '17 at 6:12

Koushik

7601117

answered Jan 24 '17 at 6:12

Koushik

7601117

answered Jan 24 '17 at 6:12

Koushik

7601117

add a comment |

In Swift 4.2 and Xcode 10.1

let string:String = "789"

let intValue:Int = Int(string)!

print(intValue)



let integerValue:Int = 789

let stringValue:String = String(integerValue)

    //OR

//let stringValue:String = "(integerValue)"

print(stringValue)

edited Jan 1 at 5:04

answered Oct 12 '18 at 6:50

iOS

2,76722047

add a comment |

In Swift 4.2 and Xcode 10.1

let string:String = "789"

let intValue:Int = Int(string)!

print(intValue)



let integerValue:Int = 789

let stringValue:String = String(integerValue)

    //OR

//let stringValue:String = "(integerValue)"

print(stringValue)

edited Jan 1 at 5:04

answered Oct 12 '18 at 6:50

iOS

2,76722047

add a comment |

In Swift 4.2 and Xcode 10.1

let string:String = "789"

let intValue:Int = Int(string)!

print(intValue)



let integerValue:Int = 789

let stringValue:String = String(integerValue)

    //OR

//let stringValue:String = "(integerValue)"

print(stringValue)

edited Jan 1 at 5:04

answered Oct 12 '18 at 6:50

iOS

2,76722047

In Swift 4.2 and Xcode 10.1

let string:String = "789"

let intValue:Int = Int(string)!

print(intValue)



let integerValue:Int = 789

let stringValue:String = String(integerValue)

    //OR

//let stringValue:String = "(integerValue)"

print(stringValue)

edited Jan 1 at 5:04

answered Oct 12 '18 at 6:50

iOS

2,76722047

edited Jan 1 at 5:04

answered Oct 12 '18 at 6:50

iOS

2,76722047

answered Oct 12 '18 at 6:50

iOS

2,76722047

answered Oct 12 '18 at 6:50

iOS

2,76722047

add a comment |

Because a string might contain non-numerical characters you should use a guard to protect the operation. Example:

guard let labelInt:Int = Int(labelString) else {

    return

}



useLabelInt()

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

add a comment |

Because a string might contain non-numerical characters you should use a guard to protect the operation. Example:

guard let labelInt:Int = Int(labelString) else {

    return

}



useLabelInt()

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

add a comment |

Because a string might contain non-numerical characters you should use a guard to protect the operation. Example:

guard let labelInt:Int = Int(labelString) else {

    return

}



useLabelInt()

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

Because a string might contain non-numerical characters you should use a guard to protect the operation. Example:

guard let labelInt:Int = Int(labelString) else {

    return

}



useLabelInt()

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

answered Feb 22 '18 at 8:38

RonTLV

1,2262923

add a comment |

Use this:

// get the values from text boxes

    let a:Double = firstText.text.bridgeToObjectiveC().doubleValue

    let b:Double = secondText.text.bridgeToObjectiveC().doubleValue



//  we checking against 0.0, because above function return 0.0 if it gets failed to convert

    if (a != 0.0) && (b != 0.0) {

        var ans = a + b

        answerLabel.text = "Answer is (ans)"

    } else {

        answerLabel.text = "Input values are not numberic"

    }

Make your UITextField KeyboardType as DecimalTab from your XIB or storyboard, and remove any if condition for doing any calculation, ie.

var ans = a + b

answerLabel.text = "Answer is (ans)"

Because keyboard type is DecimalPad there is no chance to enter other 0-9 or .

Hope this help !!

edited Jun 13 '14 at 10:03

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

add a comment |

Use this:

// get the values from text boxes

    let a:Double = firstText.text.bridgeToObjectiveC().doubleValue

    let b:Double = secondText.text.bridgeToObjectiveC().doubleValue



//  we checking against 0.0, because above function return 0.0 if it gets failed to convert

    if (a != 0.0) && (b != 0.0) {

        var ans = a + b

        answerLabel.text = "Answer is (ans)"

    } else {

        answerLabel.text = "Input values are not numberic"

    }

Make your UITextField KeyboardType as DecimalTab from your XIB or storyboard, and remove any if condition for doing any calculation, ie.

var ans = a + b

answerLabel.text = "Answer is (ans)"

Because keyboard type is DecimalPad there is no chance to enter other 0-9 or .

Hope this help !!

edited Jun 13 '14 at 10:03

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

add a comment |

Use this:

// get the values from text boxes

    let a:Double = firstText.text.bridgeToObjectiveC().doubleValue

    let b:Double = secondText.text.bridgeToObjectiveC().doubleValue



//  we checking against 0.0, because above function return 0.0 if it gets failed to convert

    if (a != 0.0) && (b != 0.0) {

        var ans = a + b

        answerLabel.text = "Answer is (ans)"

    } else {

        answerLabel.text = "Input values are not numberic"

    }

Make your UITextField KeyboardType as DecimalTab from your XIB or storyboard, and remove any if condition for doing any calculation, ie.

var ans = a + b

answerLabel.text = "Answer is (ans)"

Because keyboard type is DecimalPad there is no chance to enter other 0-9 or .

Hope this help !!

edited Jun 13 '14 at 10:03

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

Use this:

// get the values from text boxes

    let a:Double = firstText.text.bridgeToObjectiveC().doubleValue

    let b:Double = secondText.text.bridgeToObjectiveC().doubleValue



//  we checking against 0.0, because above function return 0.0 if it gets failed to convert

    if (a != 0.0) && (b != 0.0) {

        var ans = a + b

        answerLabel.text = "Answer is (ans)"

    } else {

        answerLabel.text = "Input values are not numberic"

    }

Make your UITextField KeyboardType as DecimalTab from your XIB or storyboard, and remove any if condition for doing any calculation, ie.

var ans = a + b

answerLabel.text = "Answer is (ans)"

Because keyboard type is DecimalPad there is no chance to enter other 0-9 or .

Hope this help !!

edited Jun 13 '14 at 10:03

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

edited Jun 13 '14 at 10:03

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

answered Jun 13 '14 at 6:51

Narendar Singh Saini

2,4101814

add a comment |

//  To convert user input (i.e string) to int for calculation.I did this , and it works.





    let num:Int? = Int(firstTextField.text!);



    let sum:Int = num!-2



    print(sum);

edited Nov 21 '15 at 10:43

answered Nov 20 '15 at 2:55

Jenny

277

add a comment |

//  To convert user input (i.e string) to int for calculation.I did this , and it works.





    let num:Int? = Int(firstTextField.text!);



    let sum:Int = num!-2



    print(sum);

edited Nov 21 '15 at 10:43

answered Nov 20 '15 at 2:55

Jenny

277

add a comment |

//  To convert user input (i.e string) to int for calculation.I did this , and it works.





    let num:Int? = Int(firstTextField.text!);



    let sum:Int = num!-2



    print(sum);

edited Nov 21 '15 at 10:43

answered Nov 20 '15 at 2:55

Jenny

277

//  To convert user input (i.e string) to int for calculation.I did this , and it works.





    let num:Int? = Int(firstTextField.text!);



    let sum:Int = num!-2



    print(sum);

edited Nov 21 '15 at 10:43

answered Nov 20 '15 at 2:55

Jenny

277

edited Nov 21 '15 at 10:43

answered Nov 20 '15 at 2:55

Jenny

277

answered Nov 20 '15 at 2:55

Jenny

277

answered Nov 20 '15 at 2:55

Jenny

277

add a comment |

This works for me

var a:Int? = Int(userInput.text!)

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

answered Oct 15 '15 at 21:52

Naishta

6,17934039

How is this different from the solution given in the comment?

– Prune
Oct 15 '15 at 22:25

2

The solution given in the comment is missing "!" at the end , which is expected in Swift 2 and onwards

– Naishta
Oct 22 '15 at 20:45

add a comment |

This works for me

var a:Int? = Int(userInput.text!)

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

answered Oct 15 '15 at 21:52

Naishta

6,17934039

How is this different from the solution given in the comment?

– Prune
Oct 15 '15 at 22:25

2

The solution given in the comment is missing "!" at the end , which is expected in Swift 2 and onwards

– Naishta
Oct 22 '15 at 20:45

add a comment |

This works for me

var a:Int? = Int(userInput.text!)

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

answered Oct 15 '15 at 21:52

Naishta

6,17934039

This works for me

var a:Int? = Int(userInput.text!)

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

answered Oct 15 '15 at 21:52

Naishta

6,17934039

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

edited Dec 2 '16 at 5:14

rptwsthi

8,56885393

answered Oct 15 '15 at 21:52

Naishta

6,17934039

answered Oct 15 '15 at 21:52

Naishta

6,17934039

answered Oct 15 '15 at 21:52

Naishta

6,17934039

How is this different from the solution given in the comment?

– Prune
Oct 15 '15 at 22:25

2

The solution given in the comment is missing "!" at the end , which is expected in Swift 2 and onwards

– Naishta
Oct 22 '15 at 20:45

add a comment |

How is this different from the solution given in the comment?

– Prune
Oct 15 '15 at 22:25

2

The solution given in the comment is missing "!" at the end , which is expected in Swift 2 and onwards

– Naishta
Oct 22 '15 at 20:45

How is this different from the solution given in the comment?

– Prune
Oct 15 '15 at 22:25

The solution given in the comment is missing "!" at the end , which is expected in Swift 2 and onwards

– Naishta
Oct 22 '15 at 20:45

add a comment |

for Swift3.x

extension String {

    func toInt(defaultValue: Int) -> Int {

        if let n = Int(self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)) {

            return n

        } else {

            return defaultValue

        }

    }

}

answered Apr 28 '17 at 13:02

User9527

2,7852129

add a comment |

for Swift3.x

extension String {

    func toInt(defaultValue: Int) -> Int {

        if let n = Int(self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)) {

            return n

        } else {

            return defaultValue

        }

    }

}

answered Apr 28 '17 at 13:02

User9527

2,7852129

add a comment |

for Swift3.x

extension String {

    func toInt(defaultValue: Int) -> Int {

        if let n = Int(self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)) {

            return n

        } else {

            return defaultValue

        }

    }

}

answered Apr 28 '17 at 13:02

User9527

2,7852129

for Swift3.x

extension String {

    func toInt(defaultValue: Int) -> Int {

        if let n = Int(self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)) {

            return n

        } else {

            return defaultValue

        }

    }

}

answered Apr 28 '17 at 13:02

User9527

2,7852129

answered Apr 28 '17 at 13:02

User9527

2,7852129

answered Apr 28 '17 at 13:02

User9527

2,7852129

answered Apr 28 '17 at 13:02

User9527

2,7852129

add a comment |

In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String

Int(myString)

In your case, you could use Int(textField.text!) insted of textField.text!.toInt()

Swift 1.x

let myString: String = "256"

let myInt: Int? = myString.toInt()

Swift 2.x, 3.x

let myString: String = "256"

let myInt: Int? = Int(myString)

answered Mar 16 '18 at 7:32

Azarmanesh

235

add a comment |

In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String

Int(myString)

In your case, you could use Int(textField.text!) insted of textField.text!.toInt()

Swift 1.x

let myString: String = "256"

let myInt: Int? = myString.toInt()

Swift 2.x, 3.x

let myString: String = "256"

let myInt: Int? = Int(myString)

answered Mar 16 '18 at 7:32

Azarmanesh

235

add a comment |

In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String

Int(myString)

In your case, you could use Int(textField.text!) insted of textField.text!.toInt()

Swift 1.x

let myString: String = "256"

let myInt: Int? = myString.toInt()

Swift 2.x, 3.x

let myString: String = "256"

let myInt: Int? = Int(myString)

answered Mar 16 '18 at 7:32

Azarmanesh

235

In Swift 2.x, the .toInt() function was removed from String. In replacement, Int now has an initializer that accepts a String

Int(myString)

In your case, you could use Int(textField.text!) insted of textField.text!.toInt()

Swift 1.x

let myString: String = "256"

let myInt: Int? = myString.toInt()

Swift 2.x, 3.x

let myString: String = "256"

let myInt: Int? = Int(myString)

answered Mar 16 '18 at 7:32

Azarmanesh

235

answered Mar 16 '18 at 7:32

Azarmanesh

235

answered Mar 16 '18 at 7:32

Azarmanesh

235

answered Mar 16 '18 at 7:32

Azarmanesh

235

add a comment |

for Alternative solution. You can use extension a native type. You can test with playground.

extension String {

    func add(a: Int) -> Int? {

        if let b = Int(self) {

            return b + a

        }

        else {

            return nil

        }

    }     

}

"2".add(1)

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

add a comment |

for Alternative solution. You can use extension a native type. You can test with playground.

extension String {

    func add(a: Int) -> Int? {

        if let b = Int(self) {

            return b + a

        }

        else {

            return nil

        }

    }     

}

"2".add(1)

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

add a comment |

for Alternative solution. You can use extension a native type. You can test with playground.

extension String {

    func add(a: Int) -> Int? {

        if let b = Int(self) {

            return b + a

        }

        else {

            return nil

        }

    }     

}

"2".add(1)

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

for Alternative solution. You can use extension a native type. You can test with playground.

extension String {

    func add(a: Int) -> Int? {

        if let b = Int(self) {

            return b + a

        }

        else {

            return nil

        }

    }     

}

"2".add(1)

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

answered Apr 9 '16 at 16:12

Durul Dalkanat

4,95632732

add a comment |

My solution is to have a general extension for string to int conversion.

extension String {



 // default: it is a number suitable for your project if the string is not an integer



    func toInt(default: Int) -> Int {

        if let result = Int(self) {

            return result

        }

        else {

            return default  

        }

    }



}

edited Jan 4 '17 at 11:44

answered Jan 4 '17 at 8:28

flame3

1,4581221

add a comment |

My solution is to have a general extension for string to int conversion.

extension String {



 // default: it is a number suitable for your project if the string is not an integer



    func toInt(default: Int) -> Int {

        if let result = Int(self) {

            return result

        }

        else {

            return default  

        }

    }



}

edited Jan 4 '17 at 11:44

answered Jan 4 '17 at 8:28

flame3

1,4581221

add a comment |

My solution is to have a general extension for string to int conversion.

extension String {



 // default: it is a number suitable for your project if the string is not an integer



    func toInt(default: Int) -> Int {

        if let result = Int(self) {

            return result

        }

        else {

            return default  

        }

    }



}

edited Jan 4 '17 at 11:44

answered Jan 4 '17 at 8:28

flame3

1,4581221

My solution is to have a general extension for string to int conversion.

extension String {



 // default: it is a number suitable for your project if the string is not an integer



    func toInt(default: Int) -> Int {

        if let result = Int(self) {

            return result

        }

        else {

            return default  

        }

    }



}

edited Jan 4 '17 at 11:44

answered Jan 4 '17 at 8:28

flame3

1,4581221

edited Jan 4 '17 at 11:44

answered Jan 4 '17 at 8:28

flame3

1,4581221

answered Jan 4 '17 at 8:28

flame3

1,4581221

answered Jan 4 '17 at 8:28

flame3

1,4581221

add a comment |

@IBAction func calculateAclr(_ sender: Any) {

    if let addition = addition(arrayString: [txtBox1.text, txtBox2.text, txtBox3.text]) {

      print("Answer = (addition)")

      lblAnswer.text = "(addition)"

    }

}



func addition(arrayString: [Any?]) -> Int? {



    var answer:Int?

    for arrayElement in arrayString {

        if let stringValue = arrayElement, let intValue = Int(stringValue)  {

            answer = (answer ?? 0) + intValue

        }

    }



    return answer

}

edited Jul 24 '17 at 9:59

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

add a comment |

@IBAction func calculateAclr(_ sender: Any) {

    if let addition = addition(arrayString: [txtBox1.text, txtBox2.text, txtBox3.text]) {

      print("Answer = (addition)")

      lblAnswer.text = "(addition)"

    }

}



func addition(arrayString: [Any?]) -> Int? {



    var answer:Int?

    for arrayElement in arrayString {

        if let stringValue = arrayElement, let intValue = Int(stringValue)  {

            answer = (answer ?? 0) + intValue

        }

    }



    return answer

}

edited Jul 24 '17 at 9:59

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

add a comment |

@IBAction func calculateAclr(_ sender: Any) {

    if let addition = addition(arrayString: [txtBox1.text, txtBox2.text, txtBox3.text]) {

      print("Answer = (addition)")

      lblAnswer.text = "(addition)"

    }

}



func addition(arrayString: [Any?]) -> Int? {



    var answer:Int?

    for arrayElement in arrayString {

        if let stringValue = arrayElement, let intValue = Int(stringValue)  {

            answer = (answer ?? 0) + intValue

        }

    }



    return answer

}

edited Jul 24 '17 at 9:59

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

@IBAction func calculateAclr(_ sender: Any) {

    if let addition = addition(arrayString: [txtBox1.text, txtBox2.text, txtBox3.text]) {

      print("Answer = (addition)")

      lblAnswer.text = "(addition)"

    }

}



func addition(arrayString: [Any?]) -> Int? {



    var answer:Int?

    for arrayElement in arrayString {

        if let stringValue = arrayElement, let intValue = Int(stringValue)  {

            answer = (answer ?? 0) + intValue

        }

    }



    return answer

}

edited Jul 24 '17 at 9:59

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

edited Jul 24 '17 at 9:59

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

answered Jul 23 '17 at 6:46

Krunal

40.3k20151171

add a comment |

As of swift 3, I have to force my #%@! string & int with a "!" otherwise it just doesn't work.

For example:

let prefs = UserDefaults.standard

var counter: String!

counter = prefs.string(forKey:"counter")

print("counter: (counter!)")





var counterInt = Int(counter!)

counterInt = counterInt! + 1

print("counterInt: (counterInt!)")



OUTPUT:

counter: 1

counterInt: 2

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

Can't you just do var counterInt = counter.map { Int($0) } ? Where counter whould be a String?

– Martin
Oct 19 '17 at 16:34

@Martin - No. ? makes its optional and thus adds the word "optional" to the counter string.

– Sam B
Oct 19 '17 at 22:27

IMHO, you should not force unwrap your optionals. Prefer use guard and if let statements

– Martin
Oct 20 '17 at 7:56

add a comment |

As of swift 3, I have to force my #%@! string & int with a "!" otherwise it just doesn't work.

For example:

let prefs = UserDefaults.standard

var counter: String!

counter = prefs.string(forKey:"counter")

print("counter: (counter!)")





var counterInt = Int(counter!)

counterInt = counterInt! + 1

print("counterInt: (counterInt!)")



OUTPUT:

counter: 1

counterInt: 2

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

Can't you just do var counterInt = counter.map { Int($0) } ? Where counter whould be a String?

– Martin
Oct 19 '17 at 16:34

@Martin - No. ? makes its optional and thus adds the word "optional" to the counter string.

– Sam B
Oct 19 '17 at 22:27

IMHO, you should not force unwrap your optionals. Prefer use guard and if let statements

– Martin
Oct 20 '17 at 7:56

add a comment |

As of swift 3, I have to force my #%@! string & int with a "!" otherwise it just doesn't work.

For example:

let prefs = UserDefaults.standard

var counter: String!

counter = prefs.string(forKey:"counter")

print("counter: (counter!)")





var counterInt = Int(counter!)

counterInt = counterInt! + 1

print("counterInt: (counterInt!)")



OUTPUT:

counter: 1

counterInt: 2

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

As of swift 3, I have to force my #%@! string & int with a "!" otherwise it just doesn't work.

For example:

let prefs = UserDefaults.standard

var counter: String!

counter = prefs.string(forKey:"counter")

print("counter: (counter!)")





var counterInt = Int(counter!)

counterInt = counterInt! + 1

print("counterInt: (counterInt!)")



OUTPUT:

counter: 1

counterInt: 2

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

answered Oct 16 '17 at 21:41

Sam B

19.7k1172110

Can't you just do var counterInt = counter.map { Int($0) } ? Where counter whould be a String?

– Martin
Oct 19 '17 at 16:34

@Martin - No. ? makes its optional and thus adds the word "optional" to the counter string.

– Sam B
Oct 19 '17 at 22:27

IMHO, you should not force unwrap your optionals. Prefer use guard and if let statements

– Martin
Oct 20 '17 at 7:56

add a comment |

Can't you just do var counterInt = counter.map { Int($0) } ? Where counter whould be a String?

– Martin
Oct 19 '17 at 16:34

@Martin - No. ? makes its optional and thus adds the word "optional" to the counter string.

– Sam B
Oct 19 '17 at 22:27

IMHO, you should not force unwrap your optionals. Prefer use guard and if let statements

– Martin
Oct 20 '17 at 7:56

Can't you just do var counterInt = counter.map { Int($0) } ? Where counter whould be a String?

– Martin
Oct 19 '17 at 16:34

@Martin - No. ? makes its optional and thus adds the word "optional" to the counter string.

– Sam B
Oct 19 '17 at 22:27

IMHO, you should not force unwrap your optionals. Prefer use guard and if let statements

– Martin
Oct 20 '17 at 7:56

add a comment |

Question : string "4.0000" can not be convert into integer using Int("4.000")?

Solution : NSString(string: "45.000").integerValue or Int(Float("45.000")!)! to get correct result.

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

add a comment |

Question : string "4.0000" can not be convert into integer using Int("4.000")?

Solution : NSString(string: "45.000").integerValue or Int(Float("45.000")!)! to get correct result.

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

add a comment |

Question : string "4.0000" can not be convert into integer using Int("4.000")?

Solution : NSString(string: "45.000").integerValue or Int(Float("45.000")!)! to get correct result.

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

Question : string "4.0000" can not be convert into integer using Int("4.000")?

Solution : NSString(string: "45.000").integerValue or Int(Float("45.000")!)! to get correct result.

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

answered Jan 4 '18 at 14:12

Ravi H Malviya

769518

add a comment |

I recently got the same issue. Below solution is work for me:

        let strValue = "123"

        let result = (strValue as NSString).integerValue

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

add a comment |

I recently got the same issue. Below solution is work for me:

        let strValue = "123"

        let result = (strValue as NSString).integerValue

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

add a comment |

I recently got the same issue. Below solution is work for me:

        let strValue = "123"

        let result = (strValue as NSString).integerValue

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

I recently got the same issue. Below solution is work for me:

        let strValue = "123"

        let result = (strValue as NSString).integerValue

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

answered Mar 16 '18 at 18:26

Nirmalsinh

3,27331437

add a comment |

An Int in Swift contains an initializer that accepts a String. It returns an optional Int? as the conversion can fail if the string contains not a number.

By using an if let statement you can validate whether the conversion succeeded.

So your code become something like this:

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel



@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3



    if let intAnswer = Int(txtBox1.text) {

      // Correctly converted

    }

}

answered Mar 1 at 15:50

Patrick

476313

add a comment |

An Int in Swift contains an initializer that accepts a String. It returns an optional Int? as the conversion can fail if the string contains not a number.

By using an if let statement you can validate whether the conversion succeeded.

So your code become something like this:

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel



@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3



    if let intAnswer = Int(txtBox1.text) {

      // Correctly converted

    }

}

answered Mar 1 at 15:50

Patrick

476313

add a comment |

An Int in Swift contains an initializer that accepts a String. It returns an optional Int? as the conversion can fail if the string contains not a number.

By using an if let statement you can validate whether the conversion succeeded.

So your code become something like this:

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel



@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3



    if let intAnswer = Int(txtBox1.text) {

      // Correctly converted

    }

}

answered Mar 1 at 15:50

Patrick

476313

An Int in Swift contains an initializer that accepts a String. It returns an optional Int? as the conversion can fail if the string contains not a number.

By using an if let statement you can validate whether the conversion succeeded.

So your code become something like this:

@IBOutlet var txtBox1 : UITextField

@IBOutlet var txtBox2 : UITextField

@IBOutlet var txtBox3 : UITextField

@IBOutlet var lblAnswer : UILabel



@IBAction func btn1(sender : AnyObject) {



    let answer1 = "The acceleration is"

    var answer2 = txtBox1

    var answer3 = txtBox2

    var answer4 = txtBox3



    if let intAnswer = Int(txtBox1.text) {

      // Correctly converted

    }

}

answered Mar 1 at 15:50

Patrick

476313

answered Mar 1 at 15:50

Patrick

476313

answered Mar 1 at 15:50

Patrick

476313

answered Mar 1 at 15:50

Patrick

476313

add a comment |

-1

Convert String value to Integer in Swift 4

let strValue:String = "100"

let intValue = strValue as! Int

var intValueFromString:Int = strValue as! Int

or

var intValueFromString = Int(strValue)!

edited Dec 3 '18 at 17:14

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

add a comment |

-1

Convert String value to Integer in Swift 4

let strValue:String = "100"

let intValue = strValue as! Int

var intValueFromString:Int = strValue as! Int

or

var intValueFromString = Int(strValue)!

edited Dec 3 '18 at 17:14

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

add a comment |

-1

Convert String value to Integer in Swift 4

let strValue:String = "100"

let intValue = strValue as! Int

var intValueFromString:Int = strValue as! Int

or

var intValueFromString = Int(strValue)!

edited Dec 3 '18 at 17:14

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

Convert String value to Integer in Swift 4

let strValue:String = "100"

let intValue = strValue as! Int

var intValueFromString:Int = strValue as! Int

or

var intValueFromString = Int(strValue)!

edited Dec 3 '18 at 17:14

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

edited Dec 3 '18 at 17:14

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

answered Dec 3 '18 at 16:19

Mahesh Chaudhari

18714

add a comment |

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