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java - 将第二个输入语句添加到代码后出错

问题描述

代码_1

Scanner input = new Scanner(System.in);
System.out.println("Enter Name : ");
String name = input.nextLine();
System.out.println("String name : " + name);
System.out.println("Enter age : ");
int age = input.nextInt();

代码_2

在这里，我试图同时获得姓名和年龄。此代码采用名称，在我输入名称后，它会打印错误并且不询问年龄...

System.out.println("Enter Name and age : ");
String name = input.nextLine();
int age = input.nextInt();//error may be over here

System.out.println("String name : " + name);
System.out.println("String name : " + age);

标签： javainputuser-inputnext

解决方案

问题是 nextLine 正在读取所有内容，因此 nextInt 没有可读取的内容。你想做的是。

1.读取输入

2.被某物分割（例如空格）

3.手动解析文本。

public static void main(String[] args) {
    // Try-with-resources so its closes the scanner.
    try (Scanner input = new Scanner(System.in);) {

        System.out.println("Enter Name and age : ");
        // Read input, split by spaces.
        String[] text = input.nextLine().split(" ");
        String name = "";
        // Read the values, but not the last one.
        for (int i = 0; i < text.length - 1; i++) {
            name += text[i] + " ";
        }

        try {
            // Take the last one and convert it to int.
            int age = Integer.parseInt(text[text.length - 1]);
            // Print if everything was ok.
            System.out.println("String name : " + name);
            System.out.println("String name : " + age);

        } catch (NumberFormatException e) {
            // The last field was not a number.
            System.err.println("Age has to be a number");
        }
    } // No catch propagate the exception.
}

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