使用Java扫描程序按顺序从stdin读取时引发异常

这个问题已经在这里有了答案：            >            Exception in thread “main” java.util.NoSuchElementException: No line found – Using scanner input                                     2个
我正在学习如何在Java中使用Scanner,我想从stdin中读取一个整数和一个字符串,因此我编写了以下代码：

import java.util.Scanner;

public class TryScanner {
    public void readInt() {
        Scanner in = new Scanner(System.in);
        System.out.println("Please type in an integer");
        int length = in.nextInt();
        System.out.println("Integer you type is: " + length);
        in.close();

    }
    public void readString() {
        Scanner in = new Scanner(System.in);
        System.out.println("Please type in a string");
        String s = in.nextLine();
        System.out.println("String you type is: " + s);
        in.close();

    }
    public static void main(String[] args) {
        TryScanner instance = new TryScanner();

        instance.readInt();
        // throw exception when called after readInt()
        instance.readString();
    }
}

我的readInt和readString函数可以分别很好地工作,但是,当我依次调用它们时,只有第一个函数可以工作,第二个则抛出异常：

Please type in an integer
3
Integer you type is: 3
Please type in a string
Exception in thread "main" java.util.NoSuchElementException: No line found
    at java.util.Scanner.nextLine(Scanner.java:1597)
    at helloworld.Sort.readString(Sort.java:17)
    at helloworld.Sort.main(Sort.java:27)

有人可以解释我做错了什么吗？

解决方法:

删除对

in.close();

当您关闭时,还需要全局关闭System.in(并且不会重新打开).

public void readInt() {
    Scanner in = new Scanner(System.in);
    System.out.println("Please type in an integer");
    int length = in.nextInt();
    System.out.println("Integer you type is: " + length);
    // in.close();
}
public void readString() {
    Scanner in = new Scanner(System.in);
    System.out.println("Please type in a string");
    String s = in.nextLine();
    System.out.println("String you type is: " + s);
    // in.close();
}

另外,您可以考虑将扫描仪传递给方法.

public static int readInt(Scanner in) {
    System.out.println("Please type in an integer");
    return in.nextInt();
}
public static String readString(Scanner in) {
    System.out.println("Please type in a string");
    return in.nextLine();
}
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    System.out.println("Integer you type is: " + readInt(in));
    System.out.println("String you type is: " + readString(in));
}

标签：java-util-scanner,stdin,java
来源： https://codeday.me/bug/20191028/1949983.html