How to subtract hour from DateTime in VB

[Brian Connelly]

I have a line of code which consists of two variables. One is a string called "endTime" that is formated like 05:00 PM the other is a string called "overTime" that is formated like "1". What I am trying to do is subtract the overTime from the endTime. I have the following line that converts the strings to Dates, however the subtraction returns 12:00 AM, but I am expecting 04:00 PM. Here is the code line:

timeDiff = CDate(endTime) - Hour(CDate(overTime))

Can anyone tell me what I am doing wrong and how I should go about it?

[zmbd]

Why not use the datediff() function?

http://msdn.microsoft.com/en-us/library/b5xbyt6f(v=vs.80).aspx

[Brian Connelly]

Why not use the datediff() function?
http://msdn.microsoft.com/en-us/libr...(v=vs.80).aspx

DateDiff would return the diff. in time between and one is a date the other was not. The actual solution to what I was tring to do used dateAdd(). The code is this:

timeDiff = DateAdd("h", overTime * -1, endTime)

The "h" is the metric of an hour, the overTime * -1 makes the value a negative (so Im adding a negative number to the time, and the endTime is the Time variable that I am changing.

I do appreciate your response and help.

[zmbd]

... and yet your op and the code you posted were for date data types.

So... you are after what exactly? Something along the lines of if Sam worked three hours of overtime and Sam punched off the clock at 17h00 at what time did Sam's regular time end? (answer is 14h00) If that was what you were after, and had worded your question along that lines, then my answer would have been to use the dateadd() as you have done.

[Killer42]

Probably the DateAdd function would be more appropriate.

The way it's used may vary depending on your version of VB. I've linked to the doco for VB6.