使用scanner.nextLine() [英] Using scanner.nextLine()

问题描述

我在尝试使用 java.util.Scanner 中的 nextLine() 方法时遇到了问题.

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

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:\t");
        String sentence = scanner.nextLine();

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

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

示例 1: 此示例按预期工作.行 String sentence = scanner.nextLine(); 等待输入，然后继续 System.out.print("Enter an index:\t");>.

Example #1: This example works as intended. The line String sentence = scanner.nextLine(); waits for input to be entered before continuing on to System.out.print("Enter an index:\t");.

这会产生输出:

Enter a sentence:   Hello.
Enter an index: 0

Your sentence:  Hello.
Your index: 0

<小时>

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

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

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

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

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

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

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

示例 2: 此示例未按预期工作.此示例使用 while 循环和 if - else 结构来允许用户选择要执行的操作.一旦程序到达String sentence = scanner.nextLine();，它不会等待输入，而是执行System.out.print("Enter an index:\t");.

Example #2: This example does not work as intended. This example uses a while loop and and if - else structure to allow the user to choose what to do. Once the program gets to String sentence = scanner.nextLine();, it does not wait for input but instead executes the line System.out.print("Enter an index:\t");.

这会产生输出:

Menu Options

(1) - do this
(2) - quit

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

这使得无法输入句子.

为什么示例 #2 没有按预期工作?Ex之间的唯一区别.1和2是那个Ex.2 有一个 while 循环和一个 if-else 结构.我不明白为什么这会影响scanner.nextInt() 的行为.

Why does example #2 not work as intended? The only difference between Ex. 1 and 2 is that Ex. 2 has a while loop and an if-else structure. I don't understand why this affects the behavior of scanner.nextInt().

推荐答案

我认为你的问题是

int selection = scanner.nextInt();

只读取数字，而不是行尾或数字后的任何内容.当你声明

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

String sentence = scanner.nextLine();

这会读取带有数字的行的其余部分(我怀疑的数字后面没有任何内容)

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

尝试放置一个scanner.nextLine();如果您打算忽略该行的其余部分，则在每个 nextInt() 之后.

Try placing a scanner.nextLine(); after each nextInt() if you intend to ignore the rest of the line.

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