Java的增加和时间

Question

我有四个日期，我想获得总减去...

例

timeInAM = 9：00

timeOutAM = 12：00

timeInPM = 13：00

timeOutPM = 18：00

我想让total=(timeOutAM-timeInAM)+(timeOutPM-timeInPM)这将导致总= 8：00

，但它给了我'16：00：00'

这里就是我所做的： DATE

SimpleDateFormat tf24=new SimpleDateFormat("HH:mm"); 
Date timeInAM=new Date(); 
Date timeOutAM=new Date(); 
Date timeInPM=new Date(); 
Date timeOutPM=new Date(); 
long total; 

timeInAM=tf24.parse(tblWorkPeriod.getValueAt(i, 1).toString()); 
timeOutAM=tf24.parse(tblWorkPeriod.getValueAt(i, 2).toString()); 
timeInPM=tf24.parse(tblWorkPeriod.getValueAt(i, 3).toString()); 
timeOutPM=tf24.parse(tblWorkPeriod.getValueAt(i, 4).toString()); 
total=(timeOutAM.getTime()-timeInAM.getTime())+(timeOutPM.getTime()-timeInPM.getTime()); 
System.out.println(tf24.format(new Date(total))); 

日历

Calendar timeInAM=Calendar.getInstance(); 
Calendar timeOutAM=Calendar.getInstance(); 
Calendar timeInPM=Calendar.getInstance(); 
Calendar timeOutPM=Calendar.getInstance(); 
Calendar total=Calendar.getInstance(); 

SimpleDateFormat tf24=new SimpleDateFormat("HH:mm"); 

timeInAM.setTime(tf24.parse(tblWorkPeriod.getValueAt(i, 1).toString())); 
timeOutAM.setTime(tf24.parse(tblWorkPeriod.getValueAt(i, 2).toString())); 
timeInPM.setTime(tf24.parse(tblWorkPeriod.getValueAt(i, 3).toString())); 
timeOutPM.setTime(tf24.parse(tblWorkPeriod.getValueAt(i, 4).toString())); 
long sum=(timeOutAM.getTimeInMillis()-timeInAM.getTimeInMillis())+(timeOutPM.getTimeInMillis()-timeInPM.getTimeInMillis()); 
total.setTimeInMillis(sum); 
System.out.println("total : "+tf24.format(total.getTime())); 

Answer 1

您可以试试JodaTime库（如果您可以使用其他库）。用下面你可以实现你所需要的，通过调用LocalTime::minusHours和类似的命令：

LocalTime timeInAM=new LocalTime(hourOfDay, minuteOfHour); 
LocalTime timeOutAM=new LocalTime(hourOfDay, minuteOfHour); 
LocalTime timeInPM=new LocalTime(hourOfDay, minuteOfHour); 
LocalTime timeOutPM=new LocalTime(hourOfDay, minuteOfHour); 

LocalTime amInterval = timeOutAM.minusHours(timeInAM.getHourOfDay()).minusMinutes(timeInAM.getMinuteOfHour()); 
LocalTime pmInterval = timeOutPM.minusHours(timeInPM.getHourOfDay()).minusMinutes(timeInPM.getMinuteOfHour()); 

LocalTime total = pmInterval.plusHours(amInterval.getHourOfDay()).plusMinutes(amInterval.getMinuteOfHour()); 

使用合适的DateTimeFormatter解析/打印的日期在本地时间。

Answer 2

我不敢肯定，如果我理解你的问题，但在这里：

你可以尝试使用：

double timeInAMint = Double.parseDouble(timeInAM); 
double timeOutAMint = Double.parseDouble(timeoutAM); 
double timeInPMint = Double.parseDouble(timeInPM); 
double timeOutPMint = Double.parseDouble(timeOutPM); 
double sum = timeInAMint + timeOutAMint + timeInPMint + timeOutPMint; 
System.out.println("Sum: " + sum); 

Answer 3

尝试使用Calendar实例。

编辑：示例实现。

Calendar timeInBeforeBreak = Calendar.getInstance(); 
timeInBeforeBreak.clear(); 

Calendar timeOutBeforeBreak = Calendar.getInstance(); 
timeOutBeforeBreak.clear(); 

timeInBeforeBreak.add(Calendar.HOUR_OF_DAY, 9); 
timeInBeforeBreak.add(Calendar.MINUTE, 30); 
timeOutBeforeBreak.add(Calendar.HOUR_OF_DAY, 11); 
timeOutBeforeBreak.add(Calendar.MINUTE, 30); 

long timeMillis = timeOutBeforeBreak.getTimeInMillis() - timeInBeforeBreak.getTimeInMillis(); 

System.out.println("Hours :"+timeMillis/1000/60/60);