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Access Iif Function with short date condition

by georgebee / March 5, 2007 3:53 AM PST

He everyone,

I have a query in access where I'm trying to put in a calculated field that is an Iif function. One field in the query is "Time In" which is set to a Time Format, another field is "Time Out" which is also set to a Time Format. It is basically a time log such as time in is 8:00 a.m and time out would be say 12:30 p.m.

I then have a calculated field in the query that shows how much time elapsed between the two such as - and it is formatted to SHORT TIME so it shows 4:30 based on the example above.


Now I want to make another field that based on the Iif function something like if the hours elapsed is over 5 put YES otherwise put NO if it's false..........but my problem is that I cannot get the condition over 5 so that access sees it as a short time format.

I have tried >5:00
I have tried >5

but it doesn't see it as a short time format giving me all the false returns.....I have even tried putting it enclosed in the # symbols but that only takes it as 5:00 am.....SO MY QUESTION IS, DOES ANYONE KNOW HOW TO DO THE IIF FUNCTION AND HAVE A FIELD REFERENCED IN THE CODITION BUT TO HAVE ACCESS TAKE IT AS A SHORT TIME FORMAT

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Try .2083333333333333
by Kees Bakker / March 5, 2007 5:48 PM PST

After all, 1 is a whole day = 24 hours, so 5/24 is 5 hours. Dates and times ARE numbers in MS Office, and only FORMATTED as dates.

If you really can't get it working, write a small VBA-function with the time difference as one of the parameters. Using elementary debugging, you'll find out what to write exactly. But I don't think it's necessary.

Kees

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DateDiff function
by richardp123 / March 5, 2007 6:07 PM PST

DateDiff function does the trick - look it up in help for full syntax/usage.

So DateDiff("h",TimeIn,TimeOut) would give the number of hours difference - in your example, 4.

If you made the query calculated expression:

ElapsedMins: DateDiff("n",TimeIn,TimeOut)/60

...this would give the number of minutes between the 2 time fields. In your example, this gives 270 minutes, then divided by 60 for hours which gives 4.5 hours.

Richard

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Thanks a Million
by georgebee / March 5, 2007 10:38 PM PST
In reply to: DateDiff function

Hey guys thank you so much for your replies. I did the DateDiff Function as mentioned and it works like a charm. Always knew there were the parts for the Year, Months, and DAys but didn't know to use the "N" for minutes. Thank You again for assisting me with this.

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