Question

我正在使用Java，我的程序应该要求用户输入数字1-10。如果该号码不在游侠中，则假定显示错误消息。它也应该在Pinyun中显示所选的数字。我的错误来自int number = keyboard.nextInt（）;

以下是我的其余代码：

import java.util.Scanner;
public class ForReal {


    public static void main(String[] args) {

        int number;
        Scanner keyboard = new Scanner(System.in);
        System.out.println("enter an number");
        int number = keyboard.nextInt();


        while (number < 1 || number > 10) 
            System.out.print("Error, choose a number between 1-10");

        switch (number)
        {
        case 1 : System.out.println("The English numberal 1" + number + " converts to the pinyin numeral yil1"); 
                                                    break;

        case 2:  System.out.println("The English numeral 2" + number + " converts to the pinyin numeral er4");  
                                                    break;

        case 3: System.out.println("The English numeral 3" + number + " converts to the pinyin numeral san1");
                                                    break;
        case 4: System.out.println("The English numeral 4" + number + " converts to the pinyin numeral si4");
                                                    break;
        case 5: System.out.println("The English numeral 5" + number + " converts to the pinyin numeral wu3");
                                                    break;
        case 6: System.out.println("The English numeral 6" + number + " converts to the pinyin numeral liu4");
                                                    break;
        case 7: System.out.println("The English numeral 7" + number + " converts to the pinyin numeral qil");
                                                    break;
        case 8: System.out.println("The English numeral 8" + number + " converts to the pinyin numeral bal");
                                                    break;
        case 9: System.out.println("The Arabic numeral 9" + number + " converts to the pinyin numeral jiu3");
                                                    break;
        case 10: System.out.println("The Arabic numeral 10" + number + " converts to the pinyin numeral shi2");
                                                    break;



        }


    }
}

Answer 1

您重新宣布number两次。摆脱顶部的第一个声明。这一个：

int number;

Answer 2

您尝试定义int个数字的两倍。

删除第一个声明或更改行：

int number = keyboard.nextInt();

与

number = keyboard.nextInt();

您可能还希望在try和catch中包装调用以检查来自Scanner.nextInt()方法的预期异常。

Answer 3

int number = keyboard.nextInt();
while (number < 1 || number > 10) 
    System.out.print("Error, choose a number between 1-10");

这不起作用：它只会打印消息，因为数字值在此循环内没有变化。

Answer 4

好的，首先要做的事情。

我的错误来自int number = keyboard.nextInt（）;

嗯，很高兴知道哪一行你得到错误，但什么错误会更好，只需将错误信息复制到问题，可能variable number is already defined或类似的东西

此行可能有另一个错误，如果您键入的内容不是整数，则可以使用InputMismatchException

while (number < 1 || number > 10) 
System.out.print("Error, choose a number between 1-10");

这不会改变数字的值，改为

while (number < 1 || number > 10) {
    System.out.print("Error, choose a number between 1-10");
    number = keyboard.nextInt();
}

最后：

case 1 : System.out.println("The English numberal 1" + number + " converts to the pinyin numeral yil1"); 
     break;

说真的吗？编程是懒惰人的发明者，所以可以告诉计算机做它们的工作，尽管这样的编程是可能的，尽量避免这样的假设（在更多地方使用相同的代码）。< / p>

如何清除此错误？

4 个答案: