Unexpectedly found nil while unwrapping an Optional value but value exist












0















I know I need to bind my varaible for unwrap but the problem is my value is not reconized but present.



This is my code : 



surveyW.karmaWin = Int(endedSurvey["karma"].string!)


endedSurvey is a array dictionary of my JSON backend. I get a Unexpectedly found nil while unwrapping an Optional value error.
I specify that I force the unwrapping to show you my problem.



The problem is my array contains the karma value. I show you the screen of the value:





So we can see that the value existing. Why I get a Unexpectedly found nil...?



Thanks to your reply!






swift swifty-json







share|improve this question




















  • 3





    If endedSurvey is an array you need to access the element first and then it's key endedSurvey[1] will have key value pair of "karma" not endedSurvey.

    – Mukul More
    Nov 22 '18 at 8:04






  • 2





    Possible duplicate of What does "fatal error: unexpectedly found nil while unwrapping an Optional value" mean?

    – Kamran
    Nov 22 '18 at 8:13











  • @MukulMore it is obvious from screenshot that it's a Dictionary, not an Array.

    – inokey
    Nov 22 '18 at 8:14








  • 2





    I would recommend to switch from SwiftyJSON to Decodable.

    – Sulthan
    Nov 22 '18 at 8:26






  • 1





    But if you want to keep using SwiftyJSON, at least please use it the way it's intended: .string is the optional getter, whereas .stringValue is the non-optional getter. You should not force unwrap .string yourself. Works with other types: .int vs .intValue, etc.

    – ayaio
    Nov 22 '18 at 10:07


















0















I know I need to bind my varaible for unwrap but the problem is my value is not reconized but present.



This is my code : 



surveyW.karmaWin = Int(endedSurvey["karma"].string!)


endedSurvey is a array dictionary of my JSON backend. I get a Unexpectedly found nil while unwrapping an Optional value error.
I specify that I force the unwrapping to show you my problem.



The problem is my array contains the karma value. I show you the screen of the value:





So we can see that the value existing. Why I get a Unexpectedly found nil...?



Thanks to your reply!






swift swifty-json







share|improve this question




















  • 3





    If endedSurvey is an array you need to access the element first and then it's key endedSurvey[1] will have key value pair of "karma" not endedSurvey.

    – Mukul More
    Nov 22 '18 at 8:04






  • 2





    Possible duplicate of What does "fatal error: unexpectedly found nil while unwrapping an Optional value" mean?

    – Kamran
    Nov 22 '18 at 8:13











  • @MukulMore it is obvious from screenshot that it's a Dictionary, not an Array.

    – inokey
    Nov 22 '18 at 8:14








  • 2





    I would recommend to switch from SwiftyJSON to Decodable.

    – Sulthan
    Nov 22 '18 at 8:26






  • 1





    But if you want to keep using SwiftyJSON, at least please use it the way it's intended: .string is the optional getter, whereas .stringValue is the non-optional getter. You should not force unwrap .string yourself. Works with other types: .int vs .intValue, etc.

    – ayaio
    Nov 22 '18 at 10:07
















0












0








0








I know I need to bind my varaible for unwrap but the problem is my value is not reconized but present.



This is my code : 



surveyW.karmaWin = Int(endedSurvey["karma"].string!)


endedSurvey is a array dictionary of my JSON backend. I get a Unexpectedly found nil while unwrapping an Optional value error.
I specify that I force the unwrapping to show you my problem.



The problem is my array contains the karma value. I show you the screen of the value:





So we can see that the value existing. Why I get a Unexpectedly found nil...?



Thanks to your reply!






swift swifty-json







share|improve this question
















I know I need to bind my varaible for unwrap but the problem is my value is not reconized but present.



This is my code : 



surveyW.karmaWin = Int(endedSurvey["karma"].string!)


endedSurvey is a array dictionary of my JSON backend. I get a Unexpectedly found nil while unwrapping an Optional value error.
I specify that I force the unwrapping to show you my problem.



The problem is my array contains the karma value. I show you the screen of the value:





So we can see that the value existing. Why I get a Unexpectedly found nil...?



Thanks to your reply!






swift swifty-json




swift swifty-json






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 '18 at 9:00







El-Burritos

















asked Nov 22 '18 at 8:02









El-BurritosEl-Burritos

139211




139211








  • 3





    If endedSurvey is an array you need to access the element first and then it's key endedSurvey[1] will have key value pair of "karma" not endedSurvey.

    – Mukul More
    Nov 22 '18 at 8:04






  • 2





    Possible duplicate of What does "fatal error: unexpectedly found nil while unwrapping an Optional value" mean?

    – Kamran
    Nov 22 '18 at 8:13











  • @MukulMore it is obvious from screenshot that it's a Dictionary, not an Array.

    – inokey
    Nov 22 '18 at 8:14








  • 2





    I would recommend to switch from SwiftyJSON to Decodable.

    – Sulthan
    Nov 22 '18 at 8:26






  • 1





    But if you want to keep using SwiftyJSON, at least please use it the way it's intended: .string is the optional getter, whereas .stringValue is the non-optional getter. You should not force unwrap .string yourself. Works with other types: .int vs .intValue, etc.

    – ayaio
    Nov 22 '18 at 10:07
















  • 3





    If endedSurvey is an array you need to access the element first and then it's key endedSurvey[1] will have key value pair of "karma" not endedSurvey.

    – Mukul More
    Nov 22 '18 at 8:04






  • 2





    Possible duplicate of What does "fatal error: unexpectedly found nil while unwrapping an Optional value" mean?

    – Kamran
    Nov 22 '18 at 8:13











  • @MukulMore it is obvious from screenshot that it's a Dictionary, not an Array.

    – inokey
    Nov 22 '18 at 8:14








  • 2





    I would recommend to switch from SwiftyJSON to Decodable.

    – Sulthan
    Nov 22 '18 at 8:26






  • 1





    But if you want to keep using SwiftyJSON, at least please use it the way it's intended: .string is the optional getter, whereas .stringValue is the non-optional getter. You should not force unwrap .string yourself. Works with other types: .int vs .intValue, etc.

    – ayaio
    Nov 22 '18 at 10:07










3




3





If endedSurvey is an array you need to access the element first and then it's key endedSurvey[1] will have key value pair of "karma" not endedSurvey.

– Mukul More
Nov 22 '18 at 8:04





If endedSurvey is an array you need to access the element first and then it's key endedSurvey[1] will have key value pair of "karma" not endedSurvey.

– Mukul More
Nov 22 '18 at 8:04




2




2





Possible duplicate of What does "fatal error: unexpectedly found nil while unwrapping an Optional value" mean?

– Kamran
Nov 22 '18 at 8:13





Possible duplicate of What does "fatal error: unexpectedly found nil while unwrapping an Optional value" mean?

– Kamran
Nov 22 '18 at 8:13













@MukulMore it is obvious from screenshot that it's a Dictionary, not an Array.

– inokey
Nov 22 '18 at 8:14







@MukulMore it is obvious from screenshot that it's a Dictionary, not an Array.

– inokey
Nov 22 '18 at 8:14






2




2





I would recommend to switch from SwiftyJSON to Decodable.

– Sulthan
Nov 22 '18 at 8:26





I would recommend to switch from SwiftyJSON to Decodable.

– Sulthan
Nov 22 '18 at 8:26




1




1





But if you want to keep using SwiftyJSON, at least please use it the way it's intended: .string is the optional getter, whereas .stringValue is the non-optional getter. You should not force unwrap .string yourself. Works with other types: .int vs .intValue, etc.

– ayaio
Nov 22 '18 at 10:07







But if you want to keep using SwiftyJSON, at least please use it the way it's intended: .string is the optional getter, whereas .stringValue is the non-optional getter. You should not force unwrap .string yourself. Works with other types: .int vs .intValue, etc.

– ayaio
Nov 22 '18 at 10:07














3 Answers
3






active

oldest

votes


















3














The value contained in "karma" is not String. You're trying to force cast it with SwiftyJSON but it tells you it has a nil. You first need to extract value as it is - .int, and after that convert that to something else if needed.



surveyW.karmaWin = endedSurvey["karma"].int





share|improve this answer































    0














    endedSurvey["karma"] is an Integer not string and also good way to unwrap an optional is:



    if let karma = endedSurvey["karma"] as? Int{
    
    surveyW.karmaWin = karma
    
}





    share|improve this answer
























    • This is SwiftyJSON

      – Sulthan
      Nov 22 '18 at 8:26






    • 1





      I have given a reply of native feature. Also i want to say that please don't rely on these third party libraries. Native features of apple are awesome and fast.

      – Pratyush Pratik
      Nov 22 '18 at 8:28






    • 3





      SwiftyJSON is a decent piece of software that helped a lot before Codable became available in Swift 4. Some use case are easier to write in SwiftyJSON than using Codable. Anyway, this question is about SwiftyJSON, therefore answer telling "don't use SwiftyJSON" won't be welcome.

      – Sulthan
      Nov 22 '18 at 8:31



















    0














    You can use intValue because SwiftyJSON has two kinds of "getters" for retrieving values: Optional and non-Optional



    .string and .int are the optional getters for the String and Int representation of a value, so you have to unwrap it before use




    if let fbId = fbJson["id"].string {
    
print(fbId)
    
}


    If you are 100% sure that there will always be a value, you can use the equivalent of "force unwrap" by using the non-Optional getter and you don't need if let anymore:




    let fbId = fbJson["id"].stringValue


    In your code :



    surveyW.karmaWin = endedSurvey["karma"].intValue





    share|improve this answer
























    • This work, but what's different between .int and .intValue ?

      – El-Burritos
      Nov 22 '18 at 9:08






    • 1





      .int returns an optional int, which is nil if the value is not a real int. And .intValue provides an int no matter what, which means is tries to convert other values to int values and if that's not possible then it returns a 0.

      – alireza kiani
      Nov 22 '18 at 20:52











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    The value contained in "karma" is not String. You're trying to force cast it with SwiftyJSON but it tells you it has a nil. You first need to extract value as it is - .int, and after that convert that to something else if needed.



    surveyW.karmaWin = endedSurvey["karma"].int





    share|improve this answer




























      3














      The value contained in "karma" is not String. You're trying to force cast it with SwiftyJSON but it tells you it has a nil. You first need to extract value as it is - .int, and after that convert that to something else if needed.



      surveyW.karmaWin = endedSurvey["karma"].int





      share|improve this answer


























        3












        3








        3







        The value contained in "karma" is not String. You're trying to force cast it with SwiftyJSON but it tells you it has a nil. You first need to extract value as it is - .int, and after that convert that to something else if needed.



        surveyW.karmaWin = endedSurvey["karma"].int





        share|improve this answer













        The value contained in "karma" is not String. You're trying to force cast it with SwiftyJSON but it tells you it has a nil. You first need to extract value as it is - .int, and after that convert that to something else if needed.



        surveyW.karmaWin = endedSurvey["karma"].int






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 22 '18 at 8:12









        inokeyinokey

        1,100818




        1,100818

























            0














            endedSurvey["karma"] is an Integer not string and also good way to unwrap an optional is:



            if let karma = endedSurvey["karma"] as? Int{
            
    surveyW.karmaWin = karma
            
}





            share|improve this answer
























            • This is SwiftyJSON

              – Sulthan
              Nov 22 '18 at 8:26






            • 1





              I have given a reply of native feature. Also i want to say that please don't rely on these third party libraries. Native features of apple are awesome and fast.

              – Pratyush Pratik
              Nov 22 '18 at 8:28






            • 3





              SwiftyJSON is a decent piece of software that helped a lot before Codable became available in Swift 4. Some use case are easier to write in SwiftyJSON than using Codable. Anyway, this question is about SwiftyJSON, therefore answer telling "don't use SwiftyJSON" won't be welcome.

              – Sulthan
              Nov 22 '18 at 8:31
















            0














            endedSurvey["karma"] is an Integer not string and also good way to unwrap an optional is:



            if let karma = endedSurvey["karma"] as? Int{
            
    surveyW.karmaWin = karma
            
}





            share|improve this answer
























            • This is SwiftyJSON

              – Sulthan
              Nov 22 '18 at 8:26






            • 1





              I have given a reply of native feature. Also i want to say that please don't rely on these third party libraries. Native features of apple are awesome and fast.

              – Pratyush Pratik
              Nov 22 '18 at 8:28






            • 3





              SwiftyJSON is a decent piece of software that helped a lot before Codable became available in Swift 4. Some use case are easier to write in SwiftyJSON than using Codable. Anyway, this question is about SwiftyJSON, therefore answer telling "don't use SwiftyJSON" won't be welcome.

              – Sulthan
              Nov 22 '18 at 8:31














            0












            0








            0







            endedSurvey["karma"] is an Integer not string and also good way to unwrap an optional is:



            if let karma = endedSurvey["karma"] as? Int{
            
    surveyW.karmaWin = karma
            
}





            share|improve this answer













            endedSurvey["karma"] is an Integer not string and also good way to unwrap an optional is:



            if let karma = endedSurvey["karma"] as? Int{
            
    surveyW.karmaWin = karma
            
}






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 22 '18 at 8:21









            Pratyush PratikPratyush Pratik

            511410




            511410













            • This is SwiftyJSON

              – Sulthan
              Nov 22 '18 at 8:26






            • 1





              I have given a reply of native feature. Also i want to say that please don't rely on these third party libraries. Native features of apple are awesome and fast.

              – Pratyush Pratik
              Nov 22 '18 at 8:28






            • 3





              SwiftyJSON is a decent piece of software that helped a lot before Codable became available in Swift 4. Some use case are easier to write in SwiftyJSON than using Codable. Anyway, this question is about SwiftyJSON, therefore answer telling "don't use SwiftyJSON" won't be welcome.

              – Sulthan
              Nov 22 '18 at 8:31



















            • This is SwiftyJSON

              – Sulthan
              Nov 22 '18 at 8:26






            • 1





              I have given a reply of native feature. Also i want to say that please don't rely on these third party libraries. Native features of apple are awesome and fast.

              – Pratyush Pratik
              Nov 22 '18 at 8:28






            • 3





              SwiftyJSON is a decent piece of software that helped a lot before Codable became available in Swift 4. Some use case are easier to write in SwiftyJSON than using Codable. Anyway, this question is about SwiftyJSON, therefore answer telling "don't use SwiftyJSON" won't be welcome.

              – Sulthan
              Nov 22 '18 at 8:31

















            This is SwiftyJSON

            – Sulthan
            Nov 22 '18 at 8:26





            This is SwiftyJSON

            – Sulthan
            Nov 22 '18 at 8:26




            1




            1





            I have given a reply of native feature. Also i want to say that please don't rely on these third party libraries. Native features of apple are awesome and fast.

            – Pratyush Pratik
            Nov 22 '18 at 8:28





            I have given a reply of native feature. Also i want to say that please don't rely on these third party libraries. Native features of apple are awesome and fast.

            – Pratyush Pratik
            Nov 22 '18 at 8:28




            3




            3





            SwiftyJSON is a decent piece of software that helped a lot before Codable became available in Swift 4. Some use case are easier to write in SwiftyJSON than using Codable. Anyway, this question is about SwiftyJSON, therefore answer telling "don't use SwiftyJSON" won't be welcome.

            – Sulthan
            Nov 22 '18 at 8:31





            SwiftyJSON is a decent piece of software that helped a lot before Codable became available in Swift 4. Some use case are easier to write in SwiftyJSON than using Codable. Anyway, this question is about SwiftyJSON, therefore answer telling "don't use SwiftyJSON" won't be welcome.

            – Sulthan
            Nov 22 '18 at 8:31











            0














            You can use intValue because SwiftyJSON has two kinds of "getters" for retrieving values: Optional and non-Optional



            .string and .int are the optional getters for the String and Int representation of a value, so you have to unwrap it before use




            if let fbId = fbJson["id"].string {
            
print(fbId)
            
}


            If you are 100% sure that there will always be a value, you can use the equivalent of "force unwrap" by using the non-Optional getter and you don't need if let anymore:




            let fbId = fbJson["id"].stringValue


            In your code :



            surveyW.karmaWin = endedSurvey["karma"].intValue





            share|improve this answer
























            • This work, but what's different between .int and .intValue ?

              – El-Burritos
              Nov 22 '18 at 9:08






            • 1





              .int returns an optional int, which is nil if the value is not a real int. And .intValue provides an int no matter what, which means is tries to convert other values to int values and if that's not possible then it returns a 0.

              – alireza kiani
              Nov 22 '18 at 20:52
















            0














            You can use intValue because SwiftyJSON has two kinds of "getters" for retrieving values: Optional and non-Optional



            .string and .int are the optional getters for the String and Int representation of a value, so you have to unwrap it before use




            if let fbId = fbJson["id"].string {
            
print(fbId)
            
}


            If you are 100% sure that there will always be a value, you can use the equivalent of "force unwrap" by using the non-Optional getter and you don't need if let anymore:




            let fbId = fbJson["id"].stringValue


            In your code :



            surveyW.karmaWin = endedSurvey["karma"].intValue





            share|improve this answer
























            • This work, but what's different between .int and .intValue ?

              – El-Burritos
              Nov 22 '18 at 9:08






            • 1





              .int returns an optional int, which is nil if the value is not a real int. And .intValue provides an int no matter what, which means is tries to convert other values to int values and if that's not possible then it returns a 0.

              – alireza kiani
              Nov 22 '18 at 20:52














            0












            0








            0







            You can use intValue because SwiftyJSON has two kinds of "getters" for retrieving values: Optional and non-Optional



            .string and .int are the optional getters for the String and Int representation of a value, so you have to unwrap it before use




            if let fbId = fbJson["id"].string {
            
print(fbId)
            
}


            If you are 100% sure that there will always be a value, you can use the equivalent of "force unwrap" by using the non-Optional getter and you don't need if let anymore:




            let fbId = fbJson["id"].stringValue


            In your code :



            surveyW.karmaWin = endedSurvey["karma"].intValue





            share|improve this answer













            You can use intValue because SwiftyJSON has two kinds of "getters" for retrieving values: Optional and non-Optional



            .string and .int are the optional getters for the String and Int representation of a value, so you have to unwrap it before use




            if let fbId = fbJson["id"].string {
            
print(fbId)
            
}


            If you are 100% sure that there will always be a value, you can use the equivalent of "force unwrap" by using the non-Optional getter and you don't need if let anymore:




            let fbId = fbJson["id"].stringValue


            In your code :



            surveyW.karmaWin = endedSurvey["karma"].intValue






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 22 '18 at 8:26









            alireza kianialireza kiani

            12




            12













            • This work, but what's different between .int and .intValue ?

              – El-Burritos
              Nov 22 '18 at 9:08






            • 1





              .int returns an optional int, which is nil if the value is not a real int. And .intValue provides an int no matter what, which means is tries to convert other values to int values and if that's not possible then it returns a 0.

              – alireza kiani
              Nov 22 '18 at 20:52



















            • This work, but what's different between .int and .intValue ?

              – El-Burritos
              Nov 22 '18 at 9:08






            • 1





              .int returns an optional int, which is nil if the value is not a real int. And .intValue provides an int no matter what, which means is tries to convert other values to int values and if that's not possible then it returns a 0.

              – alireza kiani
              Nov 22 '18 at 20:52

















            This work, but what's different between .int and .intValue ?

            – El-Burritos
            Nov 22 '18 at 9:08





            This work, but what's different between .int and .intValue ?

            – El-Burritos
            Nov 22 '18 at 9:08




            1




            1





            .int returns an optional int, which is nil if the value is not a real int. And .intValue provides an int no matter what, which means is tries to convert other values to int values and if that's not possible then it returns a 0.

            – alireza kiani
            Nov 22 '18 at 20:52





            .int returns an optional int, which is nil if the value is not a real int. And .intValue provides an int no matter what, which means is tries to convert other values to int values and if that's not possible then it returns a 0.

            – alireza kiani
            Nov 22 '18 at 20:52


















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            Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

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            up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
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            A Topological Invariant for $pi_3(U(n))$

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            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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