Question

我写了一个非常基本的程序，有3个选项。在执行代码的“退出”部分时，我得到一个InputMismatchError。我知道这是因为当我/用户给它一个字符串时，程序期望一个整数。我只是想知道如何设置这样的东西。我也尝试了一个字符串到char的方法，但这也给了我同样的错误。

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.Scanner;




    static Scanner S = new Scanner(System.in);
    public static void main(String[] args)  throws IOException
    {
        int DisJ, ISJ, User;

        ISJ = 1;
        DisJ = 2;
        String input = "";

        // change print outs to appropriate names::::
        System.out.println("--Main Menu--");
        System.out.println("Display Journeys:" + ISJ);
        System.out.println("Suitable Journeys:" + DisJ);
        System.out.println("Quit: " );

        //User = S.next().charAt(0);
        User = S.nextInt();
        if (User == 1)
        {

            System.out.println("You have selected Display Journeys");
            try 
            (BufferedReader ReadFile = new BufferedReader(new FileReader("E:\\input.txt"))) 
            {
                   String line = null;
                   while ((line = ReadFile.readLine()) != null) 
                   {
                   System.out.println(line);
                   }    
            }

        }

        else if (User ==2)
        {
            System.out.println("You have selected Suitable Journeys");
        }
        System.out.println("Specify Destination: ");
        String destination = S.next();
        System.out.println("Specify Max Time (HH:MM): ");
        String specificTime = S.next();
                // This assigns the first two integers to the string hours
        String hours = specificTime.substring(0,2);
                //This assigns the last two integers to the string minutes
        String minutes = specificTime.substring(3,5);
                //integer.parseInt converts the string into an integer
        int hours1 = Integer.parseInt(hours);
                //integer.parseInt converts the string into an integer
        int minutes1 = Integer.parseInt(minutes);
        int Time;
                // Equation to convert the hh:mm into minutes
        Time = (60 * hours1) + minutes1;            
        System.out.println("Specify number of changes");
        int Changes = S.nextInt();

        System.out.println( "Destination selected: " + input + "Minutes specified: "  + Time + "," + "Number of changes: " + Changes);

        int quit;
        String Quit = S.next();

        if (Quit.equals("Quit")) {

            System.out.println("Goodbye!");
            System.exit(0);

        } 
        try {
            quit = Integer.parseInt(Quit);
        }
        catch (Exception e)
        {
    System.out.println("Type Quit to leave");
        }
    }
}

Answer 1

尝试包裹yor coe并捕获错误来处理它们

int quit;
try {
    quit = Integer.parseInt(Quit);
} catch (Exception e) {
    // handle error...
}

http://beginnersbook.com/2013/04/try-catch-in-java/

简单地检查用户输入的内容只接受字符串并检查其值如下：

Scanner scanner = new Scanner(System.in);
System.out.print("Quit?");  
String text = scanner.nextLine();

if (text.equals("Quit") or text.equals("q")) {
    //...
}

Answer 2

您的问题是您无法使if语句中的参数成立吗？尝试将其设置为布尔值而不是int / string，并且在整个代码中只需将quit更改为true / false而不是数字/字符串。

编辑：哦，我明白了。所以你只想做到这一点，以便用户可以按一个键然后退出，就像“请输入Y退出程序” 你有扫描仪吗？

尝试关闭控制台时的InputMismatchError

2 个答案: