Question

Scanner issue when using nextLine after nextXXX [duplicate]

Closed 10 years ago.

Here is the skeleton of some basic code I am writing to make a simple game:

    Scanner in = new Scanner(System.in);

    String name;
    String playing;
    int age;

    do {

        System.out.println("Enter your name");
        name = in.nextLine();

        System.out.println("Enter your age");
        age = in.nextInt();

        System.out.println("Play again?");
        playing = in.nextLine();

    } while (true);

举例来说,《守则》并不像预期的那样发挥作用,这里就是《守则》的预期运作:

Enter your name
John
Enter your age
20
Play again?
Yes
Enter your name
Bill
...

然而,还有一个问题是“游戏”再行读,这是实际产出:

Enter your name
John
Enter your age
20
Play again?
Enter your name

As you can see "Enter your name" is being displayed again before "Play again?" is able to accept input. When debugging the playing variable is set to "", so there is no input that I can see and I cannot figure out what is being consumed. Any help would be appreciated, thanks!

Answer 1

<代码>nextInt( >tt 消费线末,即使int<>/code>是唯一的。

在阅读int后添加另一个nextLine(>>,要么完全放弃其价值,要么核查如果你想要阻止人们进入任何东西,那是空洞的。

Answer 2

The problem is that after calling nextInt() there is still a in the buffer and so that is what is passed. use in.next() instead of in.nextLine().

Answer 3

使用next()法,而不是使用下限法。

    Scanner in = new Scanner(System.in);

    String name;
    String playing;
    int age;

    do {

        System.out.println("Enter your name");
        name = in.next();

        System.out.println("Enter your age");
        age = in.nextInt();

        System.out.println("Play again?");
        playing = in.next();

    } while (true);

for more information refer Java Documentation for Scanner class Output :

Enter your name
Kanishka
Enter your age
23
Play again?
Yes
Enter your name
....

Answer 4

你们应当使用以下法典进行下行读。

Scanner scanner = new Scanner(System.in).useDelimiter("
");

读一下,请写

scanner.next();

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