vba access - datediff
I have 2 textboxes where the user will input a start and finish time. When I use the following code;
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
The box "Text88" will calculate the time in minutes perfectly. ie 06:00 > 14:00 = 480 minutes. However If I input say, 22:00 > 06:00, instead of equating to 480minutes it will register as -960.
How can I get it so that whatever is in the start box, it has to do up to finish, so when I input a time for 22:00 > 06:00 it will register as 480 minutes?
vba
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
add a comment |
I have 2 textboxes where the user will input a start and finish time. When I use the following code;
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
The box "Text88" will calculate the time in minutes perfectly. ie 06:00 > 14:00 = 480 minutes. However If I input say, 22:00 > 06:00, instead of equating to 480minutes it will register as -960.
How can I get it so that whatever is in the start box, it has to do up to finish, so when I input a time for 22:00 > 06:00 it will register as 480 minutes?
vba
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
add a comment |
I have 2 textboxes where the user will input a start and finish time. When I use the following code;
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
The box "Text88" will calculate the time in minutes perfectly. ie 06:00 > 14:00 = 480 minutes. However If I input say, 22:00 > 06:00, instead of equating to 480minutes it will register as -960.
How can I get it so that whatever is in the start box, it has to do up to finish, so when I input a time for 22:00 > 06:00 it will register as 480 minutes?
vba
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
I have 2 textboxes where the user will input a start and finish time. When I use the following code;
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
The box "Text88" will calculate the time in minutes perfectly. ie 06:00 > 14:00 = 480 minutes. However If I input say, 22:00 > 06:00, instead of equating to 480minutes it will register as -960.
How can I get it so that whatever is in the start box, it has to do up to finish, so when I input a time for 22:00 > 06:00 it will register as 480 minutes?
vba
vba
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
asked Nov 19 '18 at 11:18
R.LangdellR.Langdell
477
477
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You need to control when the user writes a greater time on the first textbox. Because DateDiff expects the second parameter to be a greater date than the first one. So, If is the case, before calculate the DateDiff you should apply a DateAdd to the second Time adding one day to the date. Then, do the DateDiff and the result will be what you expect.
Something like this:
If TEXTFinish > TEXTStart Then
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
Else
Text88 = DateDiff("n", TEXTStart, DateAdd("d", 1, TEXTFinish))
End If
answered Nov 19 '18 at 11:29
JortxJortx
1119
Thank you so much for your help, this works a treat!
– R.Langdell
Nov 19 '18 at 11:47
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to control when the user writes a greater time on the first textbox. Because DateDiff expects the second parameter to be a greater date than the first one. So, If is the case, before calculate the DateDiff you should apply a DateAdd to the second Time adding one day to the date. Then, do the DateDiff and the result will be what you expect.
Something like this:
If TEXTFinish > TEXTStart Then
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
Else
Text88 = DateDiff("n", TEXTStart, DateAdd("d", 1, TEXTFinish))
End If
answered Nov 19 '18 at 11:29
JortxJortx
1119
Thank you so much for your help, this works a treat!
– R.Langdell
Nov 19 '18 at 11:47
add a comment |
You need to control when the user writes a greater time on the first textbox. Because DateDiff expects the second parameter to be a greater date than the first one. So, If is the case, before calculate the DateDiff you should apply a DateAdd to the second Time adding one day to the date. Then, do the DateDiff and the result will be what you expect.
Something like this:
If TEXTFinish > TEXTStart Then
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
Else
Text88 = DateDiff("n", TEXTStart, DateAdd("d", 1, TEXTFinish))
End If
answered Nov 19 '18 at 11:29
JortxJortx
1119
Thank you so much for your help, this works a treat!
– R.Langdell
Nov 19 '18 at 11:47
add a comment |
You need to control when the user writes a greater time on the first textbox. Because DateDiff expects the second parameter to be a greater date than the first one. So, If is the case, before calculate the DateDiff you should apply a DateAdd to the second Time adding one day to the date. Then, do the DateDiff and the result will be what you expect.
Something like this:
If TEXTFinish > TEXTStart Then
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
Else
Text88 = DateDiff("n", TEXTStart, DateAdd("d", 1, TEXTFinish))
End If
answered Nov 19 '18 at 11:29
JortxJortx
1119
You need to control when the user writes a greater time on the first textbox. Because DateDiff expects the second parameter to be a greater date than the first one. So, If is the case, before calculate the DateDiff you should apply a DateAdd to the second Time adding one day to the date. Then, do the DateDiff and the result will be what you expect.
Something like this:
If TEXTFinish > TEXTStart Then
Text88 = DateDiff("n", TEXTStart, TEXTFinish)
Else
Text88 = DateDiff("n", TEXTStart, DateAdd("d", 1, TEXTFinish))
End If
answered Nov 19 '18 at 11:29
JortxJortx
1119
answered Nov 19 '18 at 11:29
JortxJortx
1119
answered Nov 19 '18 at 11:29
JortxJortx
1119
answered Nov 19 '18 at 11:29
JortxJortx
1119
1119
Thank you so much for your help, this works a treat!
– R.Langdell
Nov 19 '18 at 11:47
add a comment |
Thank you so much for your help, this works a treat!
– R.Langdell
Nov 19 '18 at 11:47
Thank you so much for your help, this works a treat!
– R.Langdell
Nov 19 '18 at 11:47
Thank you so much for your help, this works a treat!
– R.Langdell
Nov 19 '18 at 11:47
add a comment |
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