当用户在使用 Scanner 对象时输入整数时捕获异常

Question

当用户在扫描仪中输入任何整数时，我试图捕获异常。我知道使用 String item = scan.nextLine() 将允许进行任何输入，包括整数。我创建了一个自定义异常来捕获数字 1。但是如果用户输入 999 或任何其他整数会发生什么。我怎样才能捕获那些其他整数作为异常的一部分？

进口java.util.Scanner；导入 java.lang.Exception;

public class 示例一 {

public static void main(String[] args) throws TestCustomException {
    
    System.out.println("Please enter any word:");
    
    try {
        
        Scanner scan = new Scanner(System.in);
        String item = scan.nextLine();
        if (item.contains("1")) {
            throw new TestCustomException();
        } 
    }catch (TestCustomException e) {
        System.out.println("A problem occured:  " + e);
    }
    
    System.out.println("Thank you for using my application.");
    
}

}

public class TestCustomException 扩展异常 {

TestCustomException (){
    super("You can't enter any number value here");
}

}

Answer 1

正则表达式

public static void main(String[] args) {
        System.out.println("Please enter any word:");
        try {
            Scanner scan = new Scanner(System.in);
            String item = scan.nextLine();

            Pattern pattern = Pattern.compile("-?[0-9]+.?[0-9]+");
            Matcher isNum = pattern.matcher(item);

            if (isNum.matches()) {
                throw new RuntimeException();
            }
        }catch (RuntimeException e) {
            System.out.println("A problem occured:  " + e);
        }
        System.out.println("Thank you for using my application.");

    }

当用户在使用 Scanner 对象时输入整数时捕获异常

Catching an exception when a user inputs an integer while using a Scanner object

java

string

integer

exception

input