java - Java 中的两次减法

Question

在我问了一个关于减去时间的问题之前，我以为我已经知道了。然而，当我调试时，出现了一些错误。基本上我要求的是创建一个打卡钟，但我无法使用 Java Calendar API。

用户输入开始时间和打卡时间，系统会计算时间差。

我的功能不太好，我不知道为什么。例如，如果 4:00am 为开始，12:00am 为结束，则返回 8 小时 00 分钟。这是不正确的。我想我为可能性创建了一个 if 语句。

这是我的代码:

     public static void elapsedTime()
    {
        Scanner reader = new Scanner (System.in);

        System.out.print("Enter the beginning hour: ");
        int startingHour = reader.nextInt();

        System.out.print("Enter the beginning minute(s): ");
        int startingMin = reader.nextInt();


        System.out.print("Enter AM/PM: ");
        reader.nextLine();
        String startingTOD = reader.nextLine();


        System.out.print("Enter the ending hour: ");
        int endingHour = reader.nextInt();

        System.out.print("Enter the ending minute(s): ");
        int endingMin = reader.nextInt();

        System.out.print("Enter AM/PM:  ");
        reader.nextLine();
        String endingTOD = reader.nextLine();

        System.out.println();

        int hours;
        int minutes = endingMin - startingMin;
        String Am = "am" ;
        String Pm = "pm" ;

       if (endingTOD.equalsIgnoreCase(startingTOD) &&  minutes <  0  )
       {
           hours = (endingHour - startingHour) -1 ;
           minutes = minutes + 60 ;
           System.out.println(hours +  " "  + minutes);

        }
     else  if (endingHour > startingHour&& endingTOD.equalsIgnoreCase(startingTOD) &  minutes > 0  )
       {
           hours = endingHour - startingHour;
           System.out.println(hours + " " + minutes);
        }
      else  if (endingHour > startingHour && endingTOD.equalsIgnoreCase(startingTOD) && minutes == 0  )
     {
         hours = (endingHour-startingHour);
         minutes = 0;
         System.out.println(hours + " " + minutes);
     }
   else  if (endingHour < startingHour && endingTOD.equalsIgnoreCase(startingTOD) && minutes == 0  )
     {
         hours = (endingHour-startingHour) + 12;
         minutes = 0;
         System.out.println(hours + " " + minutes);
     }
   else if ( endingHour==startingHour && minutes == 0)
   {
        hours = 12;
        minutes = 0;
        System.out.println(hours + " " + minutes);
    }
    else if (( endingTOD.equalsIgnoreCase(Pm) && startingTOD.equalsIgnoreCase(Am)) && minutes > 0)
    {
        hours = (endingHour - startingHour) + 12;
        System.out.println(hours + " " + minutes);
    }

 else if  (( endingTOD.equalsIgnoreCase(Pm) && startingTOD.equalsIgnoreCase(Am)) && minutes < 0)
       {
           hours = (endingHour - startingHour) -1 ;
           minutes = minutes + 60 ;
           System.out.println(hours +  " "  + minutes);

        }
 else if (endingHour > startingHour && ( endingTOD.equalsIgnoreCase(Pm) && startingTOD.equalsIgnoreCase(Am)) && minutes == 0)
    {
        hours = (endingHour-startingHour) + 12;
        minutes = 0;
       System.out.println(hours +  " "  + minutes);
    }
    else if (endingHour < startingHour &&( endingTOD.equalsIgnoreCase(Pm) && startingTOD.equalsIgnoreCase(Am)) && minutes == 0)
    {
        hours = (endingHour - startingHour) +24;
        minutes = 0;
       System.out.println(hours +  " "  + minutes);
    }
   else if (endingHour < startingHour &&( endingTOD.equalsIgnoreCase(Pm) && startingTOD.equalsIgnoreCase(Am)) && minutes > 0)
    {
        hours = (endingHour - startingHour) + 2;
        System.out.println(hours + " " + minutes);
    }
        }

Answer 1

鉴于您无法使用 Java Calendar API，您的方法虽然不是很模块化，但乍一看似乎很合理。细节决定成败。

日期时间计算的技巧都是极端情况。这就是为什么没有人愿意编写和维护这样的东西，甚至Sun/Oracle也一次又一次地出错。

证明您的代码适用于所有条件将会很困难。您最好缩小范围并降低要求。 12 小时制还是 24 小时制？如果是 24 小时制，是从 0 点开始还是 1 点开始？我们跨越了几天吗？我们预计会穿越午夜吗？等等

但是您的任务是以 12 或 24 为模的方式比较时间(除非您的要求要求您不必“滚动”或跨越 12 或 24 小时界限。)真正的技巧是思考小时数作为一系列栅栏柱和铁轨，并计算铁轨的数量。一次性错误将成为你在这里生存的祸根。

我的建议是至少使用 Java Calendar API 来仔细检查您的结果。