Question

进行用户输入并检查用户是否输入String或int ...需要如果用户输入字符串请求再次输入整数

while(menu!=5)
        {

        System.out.println("\nWelcome To Stavan Shah's Handy Calculator");
        System.out.println("\n 1. Addition");
        System.out.println("\n 2. Subtraction");
        System.out.println("\n 3. Multiplication");
        System.out.println("\n 4. Division");
        System.out.println("\n 5. Exit");
        Scanner sc = new Scanner(System.in);        
        System.out.println("\nWhat Would you Like To Do :- ");
        c = sc.nextInt();
        switch(c)

}
public static void pauseprog(){
            System.out.println("Press Enter Key To Continue...");
            Scanner sc = new Scanner(System.in);
            sc.nextLine();
}
}

Answer 1

你可以使用这样的东西

try { 
    Integer.parseInt(userInput); 

} catch(NumberFormatException e) { 
    return false; 
}

Answer 2

这里只接受1-5：

之间输入的代码

import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;    

void checkInput() {

    Scanner s = new Scanner(System.in);
    Pattern p = Pattern.compile("^[1-5]$");  // only one digit between 1-5
    Matcher m;
    boolean match = false;

    while (!match){
        System.out.print("Enter a number between 1 and 5: ");
        m = p.matcher(s.next());
        if(m.find()){
            match = true;
            int input = m.group(0);
            // switch...
        }
    }
}

检查用户是否输入整数或字符串JAVA

2 个答案: