java - 使用scanner.nextLine（）[复制](Using scanner.nextLine() [duplicate])

Question

This question already has an answer here:

(这个问题在这里已有答案：)

• Scanner is skipping nextLine() after using next() or nextFoo()?

  (在使用next（）或nextFoo（）之后，Scanner正在跳过nextLine（）？)

  15 answers

  (15个答案)

I have been having trouble while attempting to use the nextLine() method from java.util.Scanner.

(尝试使用java.util.Scanner中的nextLine（）方法时遇到了麻烦。)

Here is what I tried:

(这是我尝试过的：)

import java.util.Scanner;

class TestRevised {
    public void menu() {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter a sentence:");
        String sentence = scanner.nextLine();

        System.out.print("Enter an index:");
        int index = scanner.nextInt();

        System.out.println("
Your sentence:" + sentence);
        System.out.println("Your index:" + index);
    }
}

Example #1: This example works as intended.

(示例＃1：此示例按预期工作。)

The line String sentence = scanner.nextLine();

(线String sentence = scanner.nextLine();)

waits for input to be entered before continuing on to System.out.print("Enter an index:\t");

(在继续执行System.out.print("Enter an index:\t");之前等待输入输入System.out.print("Enter an index:\t");)

.

(。)

This produces the output:

(这会产生输出：)

Enter a sentence:   Hello.
Enter an index: 0

Your sentence:  Hello.
Your index: 0

---

// Example #2
import java.util.Scanner;

class Test {
    public void menu() {
        Scanner scanner = new Scanner(System.in);

        while (true) {
            System.out.println("
Menu Options
");
            System.out.println("(1) - do this");
            System.out.println("(2) - quit");

            System.out.print("Please enter your selection:");
            int selection = scanner.nextInt();

            if (selection == 1) {
                System.out.print("Enter a sentence:");
                String sentence = scanner.nextLine();

                System.out.print("Enter an index:");
                int index = scanner.nextInt();

                System.out.println("
Your sentence:" + sentence);
                System.out.println("Your index:" + index);
            }
            else if (selection == 2) {
                break;
            }
        }
    }
}

Example #2: This example does not work as intended.

(示例＃2：此示例无法按预期工作。)

This example uses a while loop and and if - else structure to allow the user to choose what to do.

(此示例使用while循环和if - else结构允许用户选择要执行的操作。)

Once the program gets to String sentence = scanner.nextLine();

(一旦程序到达String sentence = scanner.nextLine();)

, it does not wait for input but instead executes the line System.out.print("Enter an index:\t");

(，它不等待输入，而是执行System.out.print("Enter an index:\t");行System.out.print("Enter an index:\t");)

.

(。)

This produces the output:

(这会产生输出：)

Menu Options

(1) - do this
(2) - quit

Please enter your selection:    1
Enter a sentence:   Enter an index: 

Which makes it impossible to enter a sentence.

(这使得无法输入句子。)

---

Why does example #2 not work as intended?

(为什么示例＃2不能按预期工作？)

The only difference between Ex.

(Ex之间的唯一区别。)

1 and 2 is that Ex.

(1和2是Ex。)

2 has a while loop and an if-else structure.

(2有一个while循环和一个if-else结构。)

I don't understand why this affects the behavior of scanner.nextInt().

(我不明白为什么这会影响scanner.nextInt（）的行为。)

  ask by Taylor P. translate from so

Answer 1

I think your problem is that

(我认为你的问题是这样的)

int selection = scanner.nextInt();

reads just the number, not the end of line or anything after the number.

(只读取数字，而不是数字后面的行尾或任何内容。)

When you declare

(当你申报时)

String sentence = scanner.nextLine();

This reads the remainder of the line with the number on it (with nothing after the number I suspect)

(这将读取该行的剩余部分（在我怀疑的数字之后没有任何内容）)

Try placing a scanner.nextLine();

(尝试放置scanner.nextLine（）;)

after each nextInt() if you intend to ignore the rest of the line.

(在每个nextInt（）之后，如果你打算忽略该行的其余部分。)