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eckler78
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Quote eckler78 Replybullet Topic: converting hours and minutes to hours
    Posted: 07 Feb 2018 at 7:50am
Hello,
I need to get the difference between two date and time fields. I have done so with the below formula, but it spits out the value in HH:MM. I need the output to be HH.HH, in hours only. So not 1:12 (one hour, 12 minutes), but 1.2 (1.2 hours).

CStr (Round (DateDiff ("n", {TA_LABOR.TIME_IN}, {TA_LABOR.TIME_OUT}) \ 60), "#") + ':' +
CStr (Round (DateDiff ("n", {TA_LABOR.TIME_IN}, {TA_LABOR.TIME_OUT}) MOD 60), "00")

Thank you for any help!

Edited by eckler78 - 08 Feb 2018 at 3:07am
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kevlray
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Quote kevlray Replybullet Posted: 07 Feb 2018 at 12:46pm
The Round (DateDiff ("n", {TA_LABOR.TIME_IN}, {TA_LABOR.TIME_OUT}) \ 60)  should get you the hours (if I am reading it correctly). So is this what you needed?
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eckler78
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Quote eckler78 Replybullet Posted: 08 Feb 2018 at 3:07am
Thank you kevlray. I have tried the formula you posted, but this returns just 1.00 hours. I need 1.2 hours. It appears to round down to the nearest hour. The formula I posted returns 1:12, one hour 12 minutes, I need this converted to 1.2 hours.
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kevlray
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Quote kevlray Replybullet Posted: 08 Feb 2018 at 4:17am
Then remove the Round function at the beginning and see what you get.
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eckler78
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Quote eckler78 Replybullet Posted: 08 Feb 2018 at 4:23am
(DateDiff ("n", {TA_LABOR.TIME_IN}, {TA_LABOR.TIME_OUT}) \ 60) still returns 1.00
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eckler78
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Quote eckler78 Replybullet Posted: 08 Feb 2018 at 4:32am
time in field = 1/22/2018 7:52:42AM
time out field = 1/22/2018 9:04:03AM
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eckler78
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Quote eckler78 Replybullet Posted: 09 Feb 2018 at 5:18am
OK, i got the below formula to work, but can anyone simplify it so it can be understood?

cstr(int((DateDiff ("n", {TA_LABOR.TIME_IN}, {TA_LABOR.TIME_OUT})) / 60),0) & "." & right(cstr((tonumber(right("0" & cstr((DateDiff ("n", {TA_LABOR.TIME_IN}, {TA_LABOR.TIME_OUT})) mod 60,0),2))/60)),2)
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