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I am using Java's Scanner to read user input. If I use nextLine only once, it works OK. With two nextLine first one doesnt wait for user to enter the string(second does).

Output:

X: Y: (wait for input)

My code

System.out.print("X: ");
x = scanner.nextLine();
System.out.print("Y: ");
y = scanner.nextLine();

Any ideas why could this happen? Thanks

user2976389
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    when i try this it works fine. y will be printed out only after user input and enter. how does your scanner look? i used Scanner scanner = new Scanner(System.in); – Sara S May 03 '14 at 22:37

1 Answers1

59

It's possible that you are calling a method like nextInt() before. Thus a program like this:

Scanner scanner = new Scanner(System.in);
int pos = scanner.nextInt();
System.out.print("X: ");
String x = scanner.nextLine();
System.out.print("Y: ");
String y = scanner.nextLine();

demonstatres the behavior you're seeing.

The problem is that nextInt() does not consume the '\n', so the next call to nextLine() consumes it and then it's waiting to read the input for y.

You need to consume the '\n' before calling nextLine(). 

System.out.print("X: ");
scanner.nextLine(); //throw away the \n not consumed by nextInt()
x = scanner.nextLine();
System.out.print("Y: ");
y = scanner.nextLine();

(actually a better way would be to call directly nextLine() after nextInt()).

Alexis C.
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    @AnubianNoob the only problem I see with this question, is that it was posted before here: http://stackoverflow.com/questions/7877529/java-string-scanner-input-does-not-wait-for-info-moves-directly-to-next-stateme?rq=1 – B537B7725DC58715F6E6BFA7AFC20C Sep 13 '15 at 03:15