Java - Scanner - Skips over my last nextLine() request

Question

So I instantiate the Scanner scan a lot earlier but it skips right over my second scan.nextLine() after scan.nextInt(). I don't understand why it skips over it?

     System.out.println("Something: ");

        String name = scan.nextLine();

        System.out.println("Something?: ");

        int number = scan.nextInt();

        System.out.println("Something?: ");

        String insurer = scan.nextLine();

        System.out.println("Something?: ");

        String another = scan.nextLine();

possible duplicate of [Skipping nextLine() after use nextInt()](http://stackoverflow.com/questions/13102045/skipping-nextline-after-use-nextint) — Am_I_Helpful, Jul 16 '14 at 01:02

score 7 · Answer 1 · answered Jul 16 '14 at 00:57

because when you enter a number

    int number = scan.nextInt();

you enter some number and hit enter, it only accepts number and keeps new line character in buffer

so nextLine() will just see the terminator character and it will assume that it is blank String as input, to fix it add one scan.nextLine() after you process int

for example:

 System.out.println("Something?: ");

 int number = scan.nextInt();

 scan.nextLine(); // <--

score 5 · Answer 2 · edited Dec 16 '19 at 07:33

The method nextInt() will not consume the new line character \n.So the new line character which was already there in the buffer before the nextInt() will be ignored.
Next when you call nextLine() after the nextInt,the nextLine() will consume the old new line
character left behind and consider the end ,skipping the rest.

enter image description here

Solution

int number = scan.nextInt();
// Adding nextLine just to discard the old \n character
scan.nextLine();

System.out.println("Something?: ");
String insurer = scan.nextLine();

OR

//Parse the string to interger explicitly

String name = scan.nextLine();
System.out.println("Something?: ");
String IntString = scanner.nextLine();

int number = Integer.valueOf(IntString);
System.out.println("Something?: ");
String insurer = scanner.nextLine();

Thank you for your detail answer! – user3590149 Jul 17 '14 at 00:48 — user3590149, Jul 17 '14 at 00:48

score 4 · Answer 3 · answered Jul 16 '14 at 00:59

4

When you call int number = scan.nextInt(); it does not consume the carriage return that has been pushed, so this is does at the next scan.nextLine();

You want your code to be

    ....
    System.out.println("Something?: ");
    int number = scan.nextInt();
    scan.nextLine();                      // add this
    System.out.println("Something?: ");
    String insurer = scan.nextLine();

answered Jul 16 '14 at 00:59

Scary Wombat

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Or quite possibly `int number = Integer.parseInt(scan.nextLine());` readInt will only read a single digit number. – LINEMAN78 Jul 16 '14 at 01:11

score 1 · Answer 4 · edited Oct 23 '17 at 14:06

All answers given before are more or less correct.

Here is a compact version:

What you want to do: First use nextInt(), then use nextLine()

What is happening: While nextInt() is waiting for your input, you press ENTER key after you type your integer. The problem is nextInt() recognizes and reads only numbers so the \n for the ENTER key is left behind on the console. When the nextLine() comes again, you expect it to wait till it finds \n. But what do you didn't see is that \n is already lying on the console because of erratic behavior of nextInt() [This problem still exists as a part of jdk8u77].

So, the nextLine reads a blank input and moves ahead.

Solution: Always add a scannerObj.nextLine() after each use of scannerObj.nextInt()

Java - Scanner - Skips over my last nextLine() request

4 Answers4

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