Question

用java写一个程序把24小时制的时间转换为12小时制的时间.具体说明内详

下面是示例对话过程:
Enter time in 24-hour notation:
13:07
That is the same as
1:07 PM
Again?(y/n)
y
Enter time in 24-hour notation:
10:15
That is the same as
10:15 AM
Again(y/n)
y
Enter time in 24-hour notation:
10:65
There is no such time as 10:65
Try again:
Enter time in 24-hour notation:
16:05
That is the same as
4:05 PM
Again?(y/n)
n
End of program

要定义一个名为TimeFormatException的异常类.如果用户输入了不合法的时间,比如10:65,甚至是无意义的东西,比如8&*68,程序将会抛出、捕获并处理一个TimeFormatException

以上就是要求,希望哪位大神能帮忙解决一下.

Answer 1

import java.util.Scanner;

public class TimeFormatException extends Exception {

    public void printException() {
        System.out.println("输入时间错误!!程序结束");
    }
    
    public TimeFormatException() {
    
    }

    public void printDate() throws TimeFormatException {
        boolean bStop = true;
        Scanner input = new Scanner(System.in);
        String reg = "([0-1][0-9]|2[0-4]):([0-5][0-9])";
        while (bStop) {
          System.out.println("请输入时间：");
          String str = input.next();
          if (str.matches(reg)) {
            int hour = Integer.parseInt(str.substring(0, 2));
            String minutes = str.substring(2, 5);
              if (hour < 12)
                System.out.println("现在时间是：" + Integer.toString(hour).concat(minutes) + " am");
              else if (hour == 12)
                System.out.println("现在时间是：" + Integer.toString(hour).concat(minutes) + " pm");
              else if(hour == 24)
                    System.out.println("现在时间是：" + "00".concat(minutes) + " am");
                else
                    System.out.println("现在时间是：" + Integer.toString(hour - 12).concat(minutes) + " pm");
          } else {
            bStop = false;
            throw new TimeFormatException();
          }
        }
     }

    public static void main(String[] args) {
        try {
            new TimeFormatException().printDate();
        } catch (TimeFormatException e) {
            e.printException();
        }
    }

}

如果看不懂 尽管问 ch_felix168_88@163.com

Answer 2

用String.splt(":");分成两部分，然后判断小时是否在0-24小时范围内，分钟也是。就这样。

追问

java基础不是很好,能麻烦你把整个代码详细写出来一下吗?

Answer 3

你看一下,这个是不是你想要的,

import java.text.SimpleDateFormat;
import java.util.Locale;
import java.util.Scanner;

public class App {

public static void main(String[] args) {

while (true) {

System.out.println("Enter time in 24-hour notation:");
Scanner sc = new Scanner(System.in);
String line = sc.nextLine();
try {
outTime(line);
} catch (TimeFormatException e) {
System.out.println("There is no such time as " + line);
System.out.println("Try again:");
continue;
}
sc = new Scanner(System.in);
line = sc.nextLine();
if ("n".equalsIgnoreCase(line)) {
break;
}

}
System.out.println("End of program");
}

public static void outTime(String line) throws TimeFormatException {
SimpleDateFormat _24time = new SimpleDateFormat("HH:mm");
SimpleDateFormat _12time = new SimpleDateFormat("hh:mm a",
Locale.ENGLISH);
try {
String[] array = line.split(":");
if (Integer.parseInt(array[0]) < 0
|| Integer.parseInt(array[0]) > 23) {
throw new TimeFormatException();
}
if (Integer.parseInt(array[1]) < 0
|| Integer.parseInt(array[1]) > 59) {
throw new TimeFormatException();
}
System.out.println(_12time.format(_24time.parse(line)));
System.out.println("Again?(y/n)");
} catch (Exception e) {
throw new TimeFormatException();
}
}

}

class TimeFormatException extends Exception {

}

Answer 4

实现这个的作用?

追问

书上的一个习题,我看了书上关于异常的内容,始终还是不明白这里如何去捕获这种异常,以及如何判断时间格式的正确与否.希望能有一个模版看看.应该就更容易明白一点了