我如何为错误的输入向用户发出错误提示？

Question

我正在尝试制作一个将温度转换为摄氏度的程序，但我卡住了如果用户尝试输入数字以外的内容，如何给用户一个错误。有人有什么建议可以帮助我吗？这是我的代码：

import java.text.DecimalFormat;
import java.util.Scanner;
public class FinalTemp {

 public static void main(String[] args) 
    
 float celsius;     
 int temp;
 String confirm = "Y/N";

 Scanner keyboard = new Scanner(System.in);


    do {
        System.out.println("Enter temperature in fahrenheit: ");
        temp = keyboard.nextInt();
    
    while (confirm.equalsIgnoreCase("N"));
    
    celsius =(float)((temp-32)*5.0/9.0);
    celsius = Float.parseFloat(new DecimalFormat("##.##").format(celsius));
    System.out.println();
    if (celsius <= 0)
        System.out.println(temp + " degrees in Fahrenheit is equal to " + celsius + " 
 degrees celsius");
    if (celsius >= 0)
        System.out.println(temp + " degrees in Fahrenheit is equal to " + celsius + " 
 degrees celsius");
    

    System.out.println("Would you like to enter another temperature?");
    System.out.println("Please type Y/N.");
    confirm = keyboard.next(); 

}
    while (confirm.equalsIgnoreCase("Y")); {
        
        System.out.println();
        System.out.println("Hello Instructor Fekete, This it the INF231 Final Assignment 
for the class");
        
}
keyboard.close();
}
}

Answer 1

可能的解决方案：

boolean invalid = true;
while (invalid) {
    try {
        String tempStr = keyboard.next();
        temp = Integer.parseInt(tempStr);
        invalid = false;
    }
    catch (Exception x) {
        System.out.println("Invalid value, inform a valid temperature");
    }
}

那么你的整个代码将是这样的：

import java.text.DecimalFormat;
import java.util.Scanner;

public class Main2 {

    public static void main(String[] args) {

        float celsius;
        int temp = 0;
        String confirm = "Y/N";

        Scanner keyboard = new Scanner(System.in);

        do {
            System.out.println("Enter temperature in fahrenheit: ");

            boolean invalid = true;
            while (invalid) {
                try {
                    String tempStr = keyboard.next();
                    temp = Integer.parseInt(tempStr);
                    invalid = false;
                }
                catch (Exception x) {
                    System.out.println("Invalid value, inform a valid temperature");
                }
            }

            while (confirm.equalsIgnoreCase("N")) ;

            celsius = (float) ((temp - 32) * 5.0 / 9.0);
            celsius = Float.parseFloat(new DecimalFormat("##.##").format(celsius));
            System.out.println();
            if (celsius <= 0)
                System.out.println(temp + " degrees in Fahrenheit is equal to " + celsius + " degrees celsius");
            if (celsius >= 0)
                System.out.println(temp + " degrees in Fahrenheit is equal to " + celsius + " degrees celsius");

                        System.out.println("Would you like to enter another temperature?");
            System.out.println("Please type Y/N.");
            confirm = keyboard.next();

        }
        while (confirm.equalsIgnoreCase("Y")); {

            System.out.println();
            System.out.println("Hello Instructor Fekete, This it the INF231 Final Assignment for the class ");

        }
        keyboard.close();
    }
}

Answer 2

您可以执行此操作的另一种方法是接受来自用户的字符串并使用 String#matches() method with a small Regular Expression （正则表达式）作为参数检查它以验证提供的是有符号或无符号整数或浮点值，例如：

String ls = System.lineSeparator();
    String confirm = "";
  
// Allow both fahrenheit & celsius to be entered as floating point values.
float temp, celsius;  
Scanner keyboard = new Scanner(System.in);
    
do {
    String temperature = "";
    while (temperature.isEmpty()) {
        System.out.print("Enter temperature in fahrenheit (c to cancel): -> ");
        temperature = keyboard.nextLine().trim();
        // Is Cancel desired?
        if (temperature.equalsIgnoreCase("c")) {
            System.out.println(ls + "Tempurture Convertion Canceled!");
            confirm = "n";
            break;
        }
        // Validate entry...
        if (!temperature.matches("-?\d+(\.\d+)?")) {
            System.out.println(ls + "Invalid Tempurature Entry (" + temperature 
                              + ")! Try again..." + ls);
            temperature = "";
        }
    }
    if (confirm.equalsIgnoreCase("n")) {
        break;
    }

    temp = Float.parseFloat(temperature);
    celsius = ((temp - 32.0f) * (5.0f / 9.0f));
    celsius = Float.parseFloat(new java.text.DecimalFormat("##.##").format(celsius));
    System.out.println();
    if (celsius <= 0.0f) {
        System.out.println(temp + " degrees Fahrenheit is equal to " 
                           + celsius + " degrees Celsius.");
    }
    else if (celsius >= 0) {
        System.out.println(temp + " degrees in Fahrenheit is equal to " 
                          + celsius + " degrees celsius");
    }
        
    while(confirm.isEmpty()) {
        System.out.println(ls + "Would you like to enter another temperature?");
        System.out.print("Please enter Y or N: -> ");
        confirm = keyboard.nextLine().trim();
        // Validate entry...
        if(!confirm.matches("(?i)[yn]")) {
            System.out.println("Invalid Responce (" + confirm + ")! Try again...");
            confirm = "";
        }
    }
    System.out.println();
} while (confirm.equalsIgnoreCase("Y")); 
    
System.out.println();
System.out.println("Hello Instructor Fekete, This it the INF231 "
                 + "Final Assignment for the class.");

使用的正则表达式：

表达式："-?\d+(\.\d+)?"访问regex101.com for explanation

表达式："(?i)[yn]"访问regex101.com for explanation

我如何为错误的输入向用户发出错误提示？

How do I give the user an error for the wrong input?

java

error-handling