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java - Trying to prompt the user to re-enter from the catch block, but the catch block terminates?

问题描述

I am trying to write a program to ask a user to enter their age and prompt them to re-enter if they enter an improper value (such as a negative number, older than 120, an age with special characters or letters, an out of range number etc...)

I tried writing a try/catch to ask the user to re-enter their age:

System.out.println("Enter your age (a positive integer): ");
    int num;

    try {
        num = in.nextInt();
        while (num < 0 || num > 120) {
            System.out.println("Bad age. Re-enter your age (a positive integer): ");
            num = in.nextInt();
        }
    } catch (InputMismatchException e) {
        //System.out.println(e);
        System.out.println("Bad age. Re-enter your age (a positive integer): ");
        num = in.nextInt();
    }

When the age entered contains special characters/letters or is out of range, the program DOES print out the words "Bad age. Re-enter your age (a positive integer)," however it immediately terminates thereafter with this error: 

Exception in thread "main" java.util.InputMismatchException
at java.base/java.util.Scanner.throwFor(Unknown Source)
at java.base/java.util.Scanner.next(Unknown Source)
at java.base/java.util.Scanner.nextInt(Unknown Source)
at java.base/java.util.Scanner.nextInt(Unknown Source)
at Age.main(Age.java:21)

My goal is to get the program to continue prompting for a valid age until the user gets its right. I would really appreciate any feedback and help. I am a java beginner :) Thank you

I tried to change put the whole code into a while loop but then it causes an infinite loop... please help!

while (num < 0 || num > 120) {
        try {
            System.out.println("Bad age. Re-enter your age (a positive integer): ");
            num = in.nextInt();
        } catch (InputMismatchException e) {
            System.out.println("Bad age. Re-enter your age (a positive integer): ");
        }
    }

标签： javaexceptionexception-handlingtry-catchinputmismatchexception

解决方案

由于您试图捕获无效的输入状态，同时仍提示用户输入正确的值，因此try-catch应将其封装在loop验证过程中。

使用 读取输入时nextInt，不会删除无效输入，因此您需要确保在尝试使用 重新读取之前清除缓冲区nextLine。或者你可以放弃它，直接读取Stringusing 的值nextLine，然后将其转换为intusing Integer.parseInt，这对个人而言比较不麻烦。

Scanner scanner = new Scanner(System.in);
int age = -1;
do {
    try {
        System.out.print("Enter ago between 0 and 250 years: ");
        String text = scanner.nextLine(); // Solves dangling new line
        age = Integer.parseInt(text);
        if (age < 0 || age > 250) {
            System.out.println("Invalid age, must be between 0 and 250");
        }
    } catch (NumberFormatException ime) {
        System.out.println("Invalid input - numbers only please");
    }
} while (age < 0 || age > 250);

使用do-while循环，基本上是因为，即使是第一次传递的有效值，您也必须至少迭代一次

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