Adding to a DateDiff

[kupe]

Can I add another clause to a DateDiff function, making the second date minus 14 days.

Rather like DateDiff("d",[OneDate], Date - 14)

please Experts?

 

[KenHigg]

Rather like DateDiff("d",[OneDate], Date() - 14)

Look like it should work.

kh

 

[kupe]

It works, but it doesn't seem to work correctly, so I thought there may be some obvious fault. Thanks and cheers, Ken.

 

[Mile-O]

DateDiff("d",[OneDate], DateAdd("d", -14, Date()))

 

[KenHigg]

What is it doing?

kh

 

[kupe]

Hi Ken
It works sometimes and then not in another place. I haven't understood why yet. But I think SJ may have the answer, which I will try.

 

[kupe]

My mistake. I was overlooking a function in the form's query that does some of the work. Thanks very much, Ken, and SJ McAbney. (I've kept your suggestions for incorporation.) Cheers

 

[kupe]

Putting the date variation into the DateDiff function works fine, like

DateDiff("d",[OneDate], Date() - 14)

But I don't doubt really I should include SJ's DateAdd. Or .. perhaps there's some unwritten rule that says if it works, leave it alone. (?) Thanks, Ken, SJ.

 

[KenHigg]

SJ's DateAdd() method is what I think I will try to use for all my date() add subtract stuff in the future as it will provide a uniform method to do most date stuff. ie: I won't be looking back at my code and wonder how I arrived a value - Did I 'DateAdd()' or did I 'date()-10'. More of a clean coding issue I guess...


kh

 

[kupe]

Yes, that's a good point. I'll follow you. Cheers