Mixing
I need to take the first N characters from the input
-2
I'm just starting on Java. I need to take the first N characters from the input. The input includes the number of characters at the begining of the input String then characters. The output includes the number at the begining.
An example, if user inputs "3stars", output would be "3st", or input "7appendices" output would be "7append".
Have tried the following:
Scanner input = new Scanner(System.in);
System.out.print("Please enter a uncoded string: ");
String first = input.nextLine();
input.close();
if (first.charAt(0) == 'u') {
first = first.toUpperCase();
} else if (first.charAt(0) == 'l') {
first = first.toLowerCase();
} else if (first.charAt(0) == 'e') {
String str = "";
for (int i = 0; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 'o') {
String str = "";
for (int i = 1; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 1++) {
String str = "";
for (int i = 1++ ; i < first.length(i); i = i charAt(i)) {
str += first.charAt(i);
}
first = str;
Any help would be great.
java numbers output
edited Nov 20 '18 at 21:09
elbraulio
748213
asked Nov 20 '18 at 20:45
WLynchWLynch
105
add a comment |
-2
I'm just starting on Java. I need to take the first N characters from the input. The input includes the number of characters at the begining of the input String then characters. The output includes the number at the begining.
An example, if user inputs "3stars", output would be "3st", or input "7appendices" output would be "7append".
Have tried the following:
Scanner input = new Scanner(System.in);
System.out.print("Please enter a uncoded string: ");
String first = input.nextLine();
input.close();
if (first.charAt(0) == 'u') {
first = first.toUpperCase();
} else if (first.charAt(0) == 'l') {
first = first.toLowerCase();
} else if (first.charAt(0) == 'e') {
String str = "";
for (int i = 0; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 'o') {
String str = "";
for (int i = 1; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 1++) {
String str = "";
for (int i = 1++ ; i < first.length(i); i = i charAt(i)) {
str += first.charAt(i);
}
first = str;
Any help would be great.
java numbers output
edited Nov 20 '18 at 21:09
elbraulio
748213
asked Nov 20 '18 at 20:45
WLynchWLynch
105
Please include the full code that you've tried rather than just this block in the middle.
– Ryan Cogswell
Nov 20 '18 at 20:55
Is last "if else" part.
– WLynch
Nov 20 '18 at 20:58
str.substring(0, str.charAt(0) - '0')
– shmosel
Nov 20 '18 at 21:01
add a comment |
-2
-2
-2
I'm just starting on Java. I need to take the first N characters from the input. The input includes the number of characters at the begining of the input String then characters. The output includes the number at the begining.
An example, if user inputs "3stars", output would be "3st", or input "7appendices" output would be "7append".
Have tried the following:
Scanner input = new Scanner(System.in);
System.out.print("Please enter a uncoded string: ");
String first = input.nextLine();
input.close();
if (first.charAt(0) == 'u') {
first = first.toUpperCase();
} else if (first.charAt(0) == 'l') {
first = first.toLowerCase();
} else if (first.charAt(0) == 'e') {
String str = "";
for (int i = 0; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 'o') {
String str = "";
for (int i = 1; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 1++) {
String str = "";
for (int i = 1++ ; i < first.length(i); i = i charAt(i)) {
str += first.charAt(i);
}
first = str;
Any help would be great.
java numbers output
edited Nov 20 '18 at 21:09
elbraulio
748213
asked Nov 20 '18 at 20:45
WLynchWLynch
105
I'm just starting on Java. I need to take the first N characters from the input. The input includes the number of characters at the begining of the input String then characters. The output includes the number at the begining.
An example, if user inputs "3stars", output would be "3st", or input "7appendices" output would be "7append".
Have tried the following:
Scanner input = new Scanner(System.in);
System.out.print("Please enter a uncoded string: ");
String first = input.nextLine();
input.close();
if (first.charAt(0) == 'u') {
first = first.toUpperCase();
} else if (first.charAt(0) == 'l') {
first = first.toLowerCase();
} else if (first.charAt(0) == 'e') {
String str = "";
for (int i = 0; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 'o') {
String str = "";
for (int i = 1; i < first.length(); i = i + 2) {
str += first.charAt(i);
}
first = str;
} else if (first.charAt(0) == 1++) {
String str = "";
for (int i = 1++ ; i < first.length(i); i = i charAt(i)) {
str += first.charAt(i);
}
first = str;
Any help would be great.
java numbers output
java numbers output
edited Nov 20 '18 at 21:09
elbraulio
748213
asked Nov 20 '18 at 20:45
WLynchWLynch
105
edited Nov 20 '18 at 21:09
elbraulio
748213
asked Nov 20 '18 at 20:45
WLynchWLynch
105
edited Nov 20 '18 at 21:09
elbraulio
748213
edited Nov 20 '18 at 21:09
elbraulio
748213
edited Nov 20 '18 at 21:09
elbraulio
748213
748213
asked Nov 20 '18 at 20:45
WLynchWLynch
105
asked Nov 20 '18 at 20:45
WLynchWLynch
105
asked Nov 20 '18 at 20:45
WLynchWLynch
105
105
Please include the full code that you've tried rather than just this block in the middle.
– Ryan Cogswell
Nov 20 '18 at 20:55
Is last "if else" part.
– WLynch
Nov 20 '18 at 20:58
str.substring(0, str.charAt(0) - '0')
– shmosel
Nov 20 '18 at 21:01
add a comment |
Please include the full code that you've tried rather than just this block in the middle.
– Ryan Cogswell
Nov 20 '18 at 20:55
Is last "if else" part.
– WLynch
Nov 20 '18 at 20:58
str.substring(0, str.charAt(0) - '0')
– shmosel
Nov 20 '18 at 21:01
Please include the full code that you've tried rather than just this block in the middle.
– Ryan Cogswell
Nov 20 '18 at 20:55
Please include the full code that you've tried rather than just this block in the middle.
– Ryan Cogswell
Nov 20 '18 at 20:55
Is last "if else" part.
– WLynch
Nov 20 '18 at 20:58
Is last "if else" part.
– WLynch
Nov 20 '18 at 20:58
str.substring(0, str.charAt(0) - '0')– shmosel
Nov 20 '18 at 21:01
str.substring(0, str.charAt(0) - '0')– shmosel
Nov 20 '18 at 21:01
add a comment |
2 Answers
2
active
oldest
votes
0
I hope this helps:
int num = Character.getNumericValue(first.charAt(0));
String str = "";
for(int i = 0;i < num;i++){
str += first.charAt(i);
}
System.out.println(str);
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
how would that be incorporated into the "else if". Just getting an error. Also is the "str += first,charAt(i)" the portion that adjusts output?
– WLynch
Nov 20 '18 at 21:08
"if user inputs "3stars", output would be "3st"" => the code i posted does this :D
– I.Stanev
Nov 20 '18 at 21:16
Just starting out, how would the final "else if" look: }
– WLynch
Nov 20 '18 at 21:20
Just "else" and put the code in :D
– I.Stanev
Nov 20 '18 at 21:24
} else int num = Character.getNumericValue(first.charAt(0)); String str = ""; for(int i = 0;i < num;i++){ str += first.charAt(i);
– WLynch
Nov 20 '18 at 21:27
|
show 3 more comments
0
this works also with input like 11abcdefghijklmnopqrstuvwxyz with numbers greater than 9
public static void main(String args) {
String input = "7appendices";
// match numbers
Pattern p = Pattern.compile("^([0-9]+)");
// by default at least the integer has one digit
int digits = 1;
int length = 0;
// while the beginning of the input match an integer
while (p.matcher(input.substring(0, digits)).matches()) {
// set length as the integer at the beginning
length = Integer.parseInt(input.substring(0, digits));
// increase the amount of digits from the integer in the input
digits++;
}
System.out.println(input.substring(0, length));
}
You only have to replace the input with scanner like this
String input = new Scanner(System.in).next();
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
I hope this helps:
int num = Character.getNumericValue(first.charAt(0));
String str = "";
for(int i = 0;i < num;i++){
str += first.charAt(i);
}
System.out.println(str);
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
how would that be incorporated into the "else if". Just getting an error. Also is the "str += first,charAt(i)" the portion that adjusts output?
– WLynch
Nov 20 '18 at 21:08
"if user inputs "3stars", output would be "3st"" => the code i posted does this :D
– I.Stanev
Nov 20 '18 at 21:16
Just starting out, how would the final "else if" look: }
– WLynch
Nov 20 '18 at 21:20
Just "else" and put the code in :D
– I.Stanev
Nov 20 '18 at 21:24
} else int num = Character.getNumericValue(first.charAt(0)); String str = ""; for(int i = 0;i < num;i++){ str += first.charAt(i);
– WLynch
Nov 20 '18 at 21:27
|
show 3 more comments
0
I hope this helps:
int num = Character.getNumericValue(first.charAt(0));
String str = "";
for(int i = 0;i < num;i++){
str += first.charAt(i);
}
System.out.println(str);
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
how would that be incorporated into the "else if". Just getting an error. Also is the "str += first,charAt(i)" the portion that adjusts output?
– WLynch
Nov 20 '18 at 21:08
"if user inputs "3stars", output would be "3st"" => the code i posted does this :D
– I.Stanev
Nov 20 '18 at 21:16
Just starting out, how would the final "else if" look: }
– WLynch
Nov 20 '18 at 21:20
Just "else" and put the code in :D
– I.Stanev
Nov 20 '18 at 21:24
} else int num = Character.getNumericValue(first.charAt(0)); String str = ""; for(int i = 0;i < num;i++){ str += first.charAt(i);
– WLynch
Nov 20 '18 at 21:27
|
show 3 more comments
0
0
0
I hope this helps:
int num = Character.getNumericValue(first.charAt(0));
String str = "";
for(int i = 0;i < num;i++){
str += first.charAt(i);
}
System.out.println(str);
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
I hope this helps:
int num = Character.getNumericValue(first.charAt(0));
String str = "";
for(int i = 0;i < num;i++){
str += first.charAt(i);
}
System.out.println(str);
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
answered Nov 20 '18 at 21:02
I.StanevI.Stanev
258
258
how would that be incorporated into the "else if". Just getting an error. Also is the "str += first,charAt(i)" the portion that adjusts output?
– WLynch
Nov 20 '18 at 21:08
"if user inputs "3stars", output would be "3st"" => the code i posted does this :D
– I.Stanev
Nov 20 '18 at 21:16
Just starting out, how would the final "else if" look: }
– WLynch
Nov 20 '18 at 21:20
Just "else" and put the code in :D
– I.Stanev
Nov 20 '18 at 21:24
} else int num = Character.getNumericValue(first.charAt(0)); String str = ""; for(int i = 0;i < num;i++){ str += first.charAt(i);
– WLynch
Nov 20 '18 at 21:27
|
show 3 more comments
how would that be incorporated into the "else if". Just getting an error. Also is the "str += first,charAt(i)" the portion that adjusts output?
– WLynch
Nov 20 '18 at 21:08
"if user inputs "3stars", output would be "3st"" => the code i posted does this :D
– I.Stanev
Nov 20 '18 at 21:16
Just starting out, how would the final "else if" look: }
– WLynch
Nov 20 '18 at 21:20
Just "else" and put the code in :D
– I.Stanev
Nov 20 '18 at 21:24
} else int num = Character.getNumericValue(first.charAt(0)); String str = ""; for(int i = 0;i < num;i++){ str += first.charAt(i);
– WLynch
Nov 20 '18 at 21:27
how would that be incorporated into the "else if". Just getting an error. Also is the "str += first,charAt(i)" the portion that adjusts output?
– WLynch
Nov 20 '18 at 21:08
how would that be incorporated into the "else if". Just getting an error. Also is the "str += first,charAt(i)" the portion that adjusts output?
– WLynch
Nov 20 '18 at 21:08
"if user inputs "3stars", output would be "3st"" => the code i posted does this :D
– I.Stanev
Nov 20 '18 at 21:16
"if user inputs "3stars", output would be "3st"" => the code i posted does this :D
– I.Stanev
Nov 20 '18 at 21:16
Just starting out, how would the final "else if" look: }
– WLynch
Nov 20 '18 at 21:20
Just starting out, how would the final "else if" look: }
– WLynch
Nov 20 '18 at 21:20
Just "else" and put the code in :D
– I.Stanev
Nov 20 '18 at 21:24
Just "else" and put the code in :D
– I.Stanev
Nov 20 '18 at 21:24
} else int num = Character.getNumericValue(first.charAt(0)); String str = ""; for(int i = 0;i < num;i++){ str += first.charAt(i);
– WLynch
Nov 20 '18 at 21:27
} else int num = Character.getNumericValue(first.charAt(0)); String str = ""; for(int i = 0;i < num;i++){ str += first.charAt(i);
– WLynch
Nov 20 '18 at 21:27
|
show 3 more comments
0
this works also with input like 11abcdefghijklmnopqrstuvwxyz with numbers greater than 9
public static void main(String args) {
String input = "7appendices";
// match numbers
Pattern p = Pattern.compile("^([0-9]+)");
// by default at least the integer has one digit
int digits = 1;
int length = 0;
// while the beginning of the input match an integer
while (p.matcher(input.substring(0, digits)).matches()) {
// set length as the integer at the beginning
length = Integer.parseInt(input.substring(0, digits));
// increase the amount of digits from the integer in the input
digits++;
}
System.out.println(input.substring(0, length));
}
You only have to replace the input with scanner like this
String input = new Scanner(System.in).next();
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
add a comment |
0
this works also with input like 11abcdefghijklmnopqrstuvwxyz with numbers greater than 9
public static void main(String args) {
String input = "7appendices";
// match numbers
Pattern p = Pattern.compile("^([0-9]+)");
// by default at least the integer has one digit
int digits = 1;
int length = 0;
// while the beginning of the input match an integer
while (p.matcher(input.substring(0, digits)).matches()) {
// set length as the integer at the beginning
length = Integer.parseInt(input.substring(0, digits));
// increase the amount of digits from the integer in the input
digits++;
}
System.out.println(input.substring(0, length));
}
You only have to replace the input with scanner like this
String input = new Scanner(System.in).next();
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
add a comment |
0
0
0
this works also with input like 11abcdefghijklmnopqrstuvwxyz with numbers greater than 9
public static void main(String args) {
String input = "7appendices";
// match numbers
Pattern p = Pattern.compile("^([0-9]+)");
// by default at least the integer has one digit
int digits = 1;
int length = 0;
// while the beginning of the input match an integer
while (p.matcher(input.substring(0, digits)).matches()) {
// set length as the integer at the beginning
length = Integer.parseInt(input.substring(0, digits));
// increase the amount of digits from the integer in the input
digits++;
}
System.out.println(input.substring(0, length));
}
You only have to replace the input with scanner like this
String input = new Scanner(System.in).next();
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
this works also with input like 11abcdefghijklmnopqrstuvwxyz with numbers greater than 9
public static void main(String args) {
String input = "7appendices";
// match numbers
Pattern p = Pattern.compile("^([0-9]+)");
// by default at least the integer has one digit
int digits = 1;
int length = 0;
// while the beginning of the input match an integer
while (p.matcher(input.substring(0, digits)).matches()) {
// set length as the integer at the beginning
length = Integer.parseInt(input.substring(0, digits));
// increase the amount of digits from the integer in the input
digits++;
}
System.out.println(input.substring(0, length));
}
You only have to replace the input with scanner like this
String input = new Scanner(System.in).next();
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
edited Nov 20 '18 at 21:51
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
answered Nov 20 '18 at 21:28
elbraulioelbraulio
748213
748213
add a comment |
add a comment |
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Please include the full code that you've tried rather than just this block in the middle.
– Ryan Cogswell
Nov 20 '18 at 20:55
Is last "if else" part.
– WLynch
Nov 20 '18 at 20:58
str.substring(0, str.charAt(0) - '0')– shmosel
Nov 20 '18 at 21:01