如何计算三次之间的差异？

Works with two different times but not showing correct calculation with three different times.

适用于两个不同的时间,但没有显示三次不同的正确计算。

SimpleDateFormat format = new SimpleDateFormat("hh:mm");
Date date1 = format.parse(timestart);
Date date2 = format.parse(timeend);
Date date3 = format.parse(breaktime);
long mills = date2.getTime() - date1.getTime() - date3.getTime();
int hours = (int)(mills/(1000 * 60 * 60));
int mins = (int) mills/(1000*60) %60;

3 个解决方案

#1

The key problem here is that date3, the duration of the break time, is using an unsuitable object. date1 and date2 are correctly defined as Date objects as they represent times of the day (10:00 and 05:00, as you mention in your comment). However, by setting date3 to be 00:30, you're declaring it to being a time of day equivalent to half past midnight today, instead you want this to be a time period of 30 minutes.

这里的关键问题是date3(中断时间的持续时间)使用的是不合适的对象。 date1和date2被正确定义为Date对象,因为它们代表当天的时间(10:00和05:00,正如您在评论中提到的那样)。但是,通过将date3设置为00:30,您宣布它是一个相当于今天午夜一半的时间,而您希望这是一个30分钟的时间段。

So you'll want to take the difference between date1 and date2, and then subtract the number of minutes taken from the break like this (assuming breakLength is 30 minutes for example):

所以你想要取date1和date2之间的差值,然后减去断点所取的分钟数(假设breakLength为30分钟):

SimpleDateFormat format = new SimpleDateFormat("hh:mm");
Date date1 = format.parse(timestart);
Date date2 = format.parse(timeend);
long breakLengthMillis = TimeUnit.MINUTES.toMillis(breakLength)

long mills = date2.getTime() - date1.getTime() - breakLengthMillis;
int hours = (int)(mills/(1000 * 60 * 60));
int mins = (int) mills/(1000*60) % 60;

Alternatively, if you want to measure a break by its start and end times, you could use this:

或者,如果要根据开始和结束时间来衡量中断,可以使用:

SimpleDateFormat format = new SimpleDateFormat("hh:mm");
Date date1 = format.parse(timestart);
Date date2 = format.parse(timeend);
Date date3 = format.parse(breakStart);       // Start time of break
Date date4 = format.parse(breakEnd);         // End time of break

long mills = date2.getTime() - date1.getTime() - (date4.getTime() - date3.getTime());
int hours = (int)(mills/(1000 * 60 * 60));
int mins = (int) mills/(1000*60) % 60;

#2

Are you trying to calculate:

你想要计算:

Work_Duration = End_Time - Start_Time - Break_Duration?

If yes, you can simply use an integer/long variable to represent break duration in units of your choice (seconds or minutes). It need not be a Date object.

如果是,您可以简单地使用整数/长变量来表示以您选择的单位(秒或分钟)为单位的中断持续时间。它不必是Date对象。

You can calculate:

你可以计算:

Break_Duration = Break_End_Time - Break_Start_Time

So, you will have 2 Date objects (one for start of break, another for the end of break time), instead of just one for breaktime.

因此,您将拥有2个Date对象(一个用于开始休息,另一个用于休息时间结束),而不是一个用于休息时间。

#3

If you want calculate the different between timeend end timestart and the different between timeend end breaktime time you need do:

如果你想要计算timeend end timestart和timeend end breaktime time之间的差异你需要做的事情:

SimpleDateFormat format = new SimpleDateFormat("hh:mm");
Date date1 = format.parse(timestart);
Date date2 = format.parse(timeend);
Date date3 = format.parse(breaktime);
long mills = date2.getTime() - date1.getTime();
long mills2 = date2.getTime() - date3.getTime();
int hours = (int)(mills/(1000 * 60 * 60));
int mins = (int) mills/(1000*60);
int hours2 = (int)(mills/(1000 * 60 * 60));
int mins2 = (int) mills/(1000*60);

Because if you do date1- date2- date3 you have a negative number (es 1535983083-1535810283-1535896683)

因为如果你做date1- date2- date3你有一个负数(es 1535983083-1535810283-1535896683)

#1