Subtracting Time and Date in hours & minutes

[gvillepa]

Hi, I have 2 columns. Each column contains a date & time in "m/d/yyyy h:mm AM/PM" format.

I am trying to subtract the differences between two columns and am having a hard time. Any assistance appreciated and the desired result format is negotiable, so long as i get minutes difference....3 hours could be 180 minutes, etc. Thanks!

[TABLE="width: 500"]
<tbody>[TR]
[TD]TIME 1[/TD]
[TD]Time 2[/TD]
[TD]Desired Result[/TD]
[/TR]
[TR]
[TD]3/4/2019 10:47 AM[/TD]
[TD]3/4/2019 10:45 AM[/TD]
[TD]-2 Minutes[/TD]
[/TR]
[TR]
[TD]3/5/2019 11:55 AM[/TD]
[TD]3/5/2019 12:55 PM[/TD]
[TD]60 minutes or 1 hour[/TD]
[/TR]
[TR]
[TD]3/6/2019 11:45 PM[/TD]
[TD]3/7/2019 12:01 AM[/TD]
[TD]16 minutes[/TD]
[/TR]
</tbody>[/TABLE]

 

[Rick Rothstein]

You can use this formula...

=MAX(A1,B1)-MIN(A1,B1)

to return the time differential. Since this return value will be a real time serial value, you can format the cell using any time format pattern that you like.

 

[gvillepa]

Thank you. I just put this together:

=SUM((E3-F3)*86400)/60

which works, but not as well as what you just offered. Thank you kind sir.

 

[CyrusTheVirus]

Clever Rick.

gvillepa, you could try something like the below.

IF(B2 < A2,(MAX(A2,B2) - MIN(A2,B2)) * - 1440,(MAX(A2,B2) - MIN(A2,B2)) * 1440)<a2,(max(a2,b2)-min(a2,b2))*-1440,(max(a2,b2)-min(a2,b2))*1440)<a2,(max(a2,b2)-min(a2,b2))*-1440,(max(a2,b2)-min(a2,b2))*1440)[ code]
<a2,(max(a2,b2)-min(a2,b2))*-1440,(max(a2,b2)-min(a2,b2))*1440)[ code]<="" html=""></a2,(max(a2,b2)-min(a2,b2))*-1440,(max(a2,b2)-min(a2,b2))*1440)[></a2,(max(a2,b2)-min(a2,b2))*-1440,(max(a2,b2)-min(a2,b2))*1440)<a2,(max(a2,b2)-min(a2,b2))*-1440,(max(a2,b2)-min(a2,b2))*1440)[>