Calculate time based on an integer in another field

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I am trying to generate a field in a table that is a time (e.g.,8:15 AM).
This time is to be calculated from another field which contains an integer.
I want the time generated to be 7:30 AM plus 7.5 minutes times the number in
the other field (group number). Can anyone tell me how to accomplish this?

Thanks for all help.
 
Hi Arnold,

I assume it's a calculated field you have in mind; there's no point
storing data that can easily be calculated whenever its needed.

If the time is stored in a date/time field, use something like this as a
calculated field in a query, or as the data source of a textbox on a
form or report:

DateAdd("n", 7.5 * [GroupNumber], [TimeField])

If the time is stored in text field, try

DateAdd("n", 7.5 * [GroupNumber], CDate([TimeField]))
 
John,

Thank you so much. That works perfectly. When I first looked at your reply
I was going to write back and ask what the "n" meant. Then I tried to look
up the adddate function in help. Unfortunately, the only thing there is an
example and there is no explanation of the function's syntax or effect. I
surmised that the n meant minutes. Can you point me to someplace that
explains the function? Your help is much appreciated.

Thanks again.

John Nurick said:
Hi Arnold,

I assume it's a calculated field you have in mind; there's no point
storing data that can easily be calculated whenever its needed.

If the time is stored in a date/time field, use something like this as a
calculated field in a query, or as the data source of a textbox on a
form or report:

DateAdd("n", 7.5 * [GroupNumber], [TimeField])

If the time is stored in text field, try

DateAdd("n", 7.5 * [GroupNumber], CDate([TimeField]))


I am trying to generate a field in a table that is a time (e.g.,8:15 AM).
This time is to be calculated from another field which contains an integer.
I want the time generated to be 7:30 AM plus 7.5 minutes times the number in
the other field (group number). Can anyone tell me how to accomplish this?

Thanks for all help.
Click to expand...
Click to expand...
 
Even though this reference says it's for Access 97, it applies to all
versions:

http://msdn.microsoft.com/archive/en-us/office97/html/output/F1/d6/S5B206.asp

--
Doug Steele, Microsoft Access MVP

(no e-mails, please!)



Arnold R said:
John,

Thank you so much. That works perfectly. When I first looked at your
reply
I was going to write back and ask what the "n" meant. Then I tried to
look
up the adddate function in help. Unfortunately, the only thing there is
an
example and there is no explanation of the function's syntax or effect. I
surmised that the n meant minutes. Can you point me to someplace that
explains the function? Your help is much appreciated.

Thanks again.

John Nurick said:
Hi Arnold,

I assume it's a calculated field you have in mind; there's no point
storing data that can easily be calculated whenever its needed.

If the time is stored in a date/time field, use something like this as a
calculated field in a query, or as the data source of a textbox on a
form or report:

DateAdd("n", 7.5 * [GroupNumber], [TimeField])

If the time is stored in text field, try

DateAdd("n", 7.5 * [GroupNumber], CDate([TimeField]))


I am trying to generate a field in a table that is a time (e.g.,8:15
AM).
This time is to be calculated from another field which contains an
integer.
I want the time generated to be 7:30 AM plus 7.5 minutes times the
number in
the other field (group number). Can anyone tell me how to accomplish
this?

Thanks for all help.
Click to expand...
Click to expand...
Click to expand...
 
Thanks very much.

Douglas J. Steele said:
Even though this reference says it's for Access 97, it applies to all
versions:

http://msdn.microsoft.com/archive/en-us/office97/html/output/F1/d6/S5B206.asp

--
Doug Steele, Microsoft Access MVP

(no e-mails, please!)



Arnold R said:
John,

Thank you so much. That works perfectly. When I first looked at your
reply
I was going to write back and ask what the "n" meant. Then I tried to
look
up the adddate function in help. Unfortunately, the only thing there is
an
example and there is no explanation of the function's syntax or effect. I
surmised that the n meant minutes. Can you point me to someplace that
explains the function? Your help is much appreciated.

Thanks again.

John Nurick said:
Hi Arnold,

I assume it's a calculated field you have in mind; there's no point
storing data that can easily be calculated whenever its needed.

If the time is stored in a date/time field, use something like this as a
calculated field in a query, or as the data source of a textbox on a
form or report:

DateAdd("n", 7.5 * [GroupNumber], [TimeField])

If the time is stored in text field, try

DateAdd("n", 7.5 * [GroupNumber], CDate([TimeField]))


On Fri, 13 May 2005 21:40:01 -0700, "Arnold R"

I am trying to generate a field in a table that is a time (e.g.,8:15
AM).
This time is to be calculated from another field which contains an
integer.
I want the time generated to be 7:30 AM plus 7.5 minutes times the
number in
the other field (group number). Can anyone tell me how to accomplish
this?

Thanks for all help.
Click to expand...
Click to expand...
Click to expand...
 
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