Mixing
Remove lunch time from Total Time in Access.
I am doing a datediff in an access field that I now need to also remove a 50 minute lunch break from. Everything I have tried has created crazy numbers. This is the function that works correctly without removing the lunch break:
=(DateDiff("n",[DispatchTime],[ReturnTime]))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime])) Mod 60,":00")
Then I try:
=(DateDiff("n",[DispatchTime],[ReturnTime]-.5))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime]-.5)) Mod 60,":00")
How can I do this? and still show in h:mm format?
ms-access
asked Nov 19 '18 at 15:02
djblois
361623
add a comment |
I am doing a datediff in an access field that I now need to also remove a 50 minute lunch break from. Everything I have tried has created crazy numbers. This is the function that works correctly without removing the lunch break:
=(DateDiff("n",[DispatchTime],[ReturnTime]))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime])) Mod 60,":00")
Then I try:
=(DateDiff("n",[DispatchTime],[ReturnTime]-.5))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime]-.5)) Mod 60,":00")
How can I do this? and still show in h:mm format?
ms-access
asked Nov 19 '18 at 15:02
djblois
361623
add a comment |
I am doing a datediff in an access field that I now need to also remove a 50 minute lunch break from. Everything I have tried has created crazy numbers. This is the function that works correctly without removing the lunch break:
=(DateDiff("n",[DispatchTime],[ReturnTime]))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime])) Mod 60,":00")
Then I try:
=(DateDiff("n",[DispatchTime],[ReturnTime]-.5))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime]-.5)) Mod 60,":00")
How can I do this? and still show in h:mm format?
ms-access
asked Nov 19 '18 at 15:02
djblois
361623
I am doing a datediff in an access field that I now need to also remove a 50 minute lunch break from. Everything I have tried has created crazy numbers. This is the function that works correctly without removing the lunch break:
=(DateDiff("n",[DispatchTime],[ReturnTime]))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime])) Mod 60,":00")
Then I try:
=(DateDiff("n",[DispatchTime],[ReturnTime]-.5))60 & Format((DateDiff("n",[DispatchTime],[ReturnTime]-.5)) Mod 60,":00")
How can I do this? and still show in h:mm format?
ms-access
ms-access
asked Nov 19 '18 at 15:02
djblois
361623
asked Nov 19 '18 at 15:02
djblois
361623
asked Nov 19 '18 at 15:02
djblois
361623
asked Nov 19 '18 at 15:02
djblois
361623
asked Nov 19 '18 at 15:02
djblois
361623
361623
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The following should suffice:
=Format(DateAdd("n",-50,[ReturnTime])-[DispatchTime],"h:nn")
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
1
And the format should be"h:nn".
– Gustav
Nov 19 '18 at 16:14
1
Thanks @Gustav - I have updated my answer above.
– Lee Mac
Nov 19 '18 at 16:25
1
Great - though thathdidn't postpone my upvote.
– Gustav
Nov 19 '18 at 16:37
add a comment |
Return Time - .5 will remove 1/2 a day e.g.12 hours.
DateAdd ("n",-50, [YourDatefield]) will subtract 50 minutes.
answered Nov 19 '18 at 15:17
Minty
1,4821411
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The following should suffice:
=Format(DateAdd("n",-50,[ReturnTime])-[DispatchTime],"h:nn")
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
1
And the format should be"h:nn".
– Gustav
Nov 19 '18 at 16:14
1
Thanks @Gustav - I have updated my answer above.
– Lee Mac
Nov 19 '18 at 16:25
1
Great - though thathdidn't postpone my upvote.
– Gustav
Nov 19 '18 at 16:37
add a comment |
The following should suffice:
=Format(DateAdd("n",-50,[ReturnTime])-[DispatchTime],"h:nn")
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
1
And the format should be"h:nn".
– Gustav
Nov 19 '18 at 16:14
1
Thanks @Gustav - I have updated my answer above.
– Lee Mac
Nov 19 '18 at 16:25
1
Great - though thathdidn't postpone my upvote.
– Gustav
Nov 19 '18 at 16:37
add a comment |
The following should suffice:
=Format(DateAdd("n",-50,[ReturnTime])-[DispatchTime],"h:nn")
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
The following should suffice:
=Format(DateAdd("n",-50,[ReturnTime])-[DispatchTime],"h:nn")
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
edited Nov 19 '18 at 16:24
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
answered Nov 19 '18 at 15:20
Lee Mac
3,50531339
3,50531339
1
And the format should be"h:nn".
– Gustav
Nov 19 '18 at 16:14
1
Thanks @Gustav - I have updated my answer above.
– Lee Mac
Nov 19 '18 at 16:25
1
Great - though thathdidn't postpone my upvote.
– Gustav
Nov 19 '18 at 16:37
add a comment |
1
And the format should be"h:nn".
– Gustav
Nov 19 '18 at 16:14
1
Thanks @Gustav - I have updated my answer above.
– Lee Mac
Nov 19 '18 at 16:25
1
Great - though thathdidn't postpone my upvote.
– Gustav
Nov 19 '18 at 16:37
1
1
And the format should be
"h:nn".– Gustav
Nov 19 '18 at 16:14
And the format should be
"h:nn".– Gustav
Nov 19 '18 at 16:14
1
1
Thanks @Gustav - I have updated my answer above.
– Lee Mac
Nov 19 '18 at 16:25
Thanks @Gustav - I have updated my answer above.
– Lee Mac
Nov 19 '18 at 16:25
1
1
Great - though that
h didn't postpone my upvote.– Gustav
Nov 19 '18 at 16:37
Great - though that
h didn't postpone my upvote.– Gustav
Nov 19 '18 at 16:37
add a comment |
Return Time - .5 will remove 1/2 a day e.g.12 hours.
DateAdd ("n",-50, [YourDatefield]) will subtract 50 minutes.
answered Nov 19 '18 at 15:17
Minty
1,4821411
add a comment |
Return Time - .5 will remove 1/2 a day e.g.12 hours.
DateAdd ("n",-50, [YourDatefield]) will subtract 50 minutes.
answered Nov 19 '18 at 15:17
Minty
1,4821411
add a comment |
Return Time - .5 will remove 1/2 a day e.g.12 hours.
DateAdd ("n",-50, [YourDatefield]) will subtract 50 minutes.
answered Nov 19 '18 at 15:17
Minty
1,4821411
Return Time - .5 will remove 1/2 a day e.g.12 hours.
DateAdd ("n",-50, [YourDatefield]) will subtract 50 minutes.
answered Nov 19 '18 at 15:17
Minty
1,4821411
answered Nov 19 '18 at 15:17
Minty
1,4821411
answered Nov 19 '18 at 15:17
Minty
1,4821411
answered Nov 19 '18 at 15:17
Minty
1,4821411
1,4821411
add a comment |
add a comment |
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