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Getting First and Last Day of Week

Posted on 2001-07-26
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Last Modified: 2012-05-04
I am creating a report where the user wants it to print the records for the current week. They asked for a similiar report except for just the current date for that one I used Date() in the query.
I would like to do something similiar in the query except getting the first and last day of the week something like

DateField1 between FirstDate and LastDate


Thanks
Ross
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Question by:bozo7
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Author Comment

by:bozo7
ID: 6323849
Oh Yeah I am using Access 2000
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Expert Comment

by:dovholuk
ID: 6323885
use the following:
first day of the week (Monday) would be: date - weekday(now, vbMonday) + 1
last day of week (Friday): (6 - weekday(now, vbMonday)) + date

use those in your between statement (not sure on the between syntax):
datefield1 between (date - weekday(now, vbMonday) + 1) and ((6 - weekday(now, vbMonday)) + date)

dovholuk
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Expert Comment

by:dovholuk
ID: 6323890
date - weekday(now, vbMonday) + 1 produces a result of 7/23/01 (monday july 23rd)
(6 - weekday(now, vbMonday)) + date yields a result of 7/28/01 (friday, july 28th)

depending on when you want your week to start and end, you can change the vbMonday in the first expression, or the 6 in the second expression.

need more help?

dovholuk
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Author Comment

by:bozo7
ID: 6323910
Thank You I will try this tonight and let you know.
Ross
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Expert Comment

by:DennisBorg
ID: 6324084
>((6 - weekday(now, vbMonday)) + date)

This seems to overshoot Friday by one day and actually targets Saturday. (July 28 is Saturday, not Friday)

Personally, I find it the following more simple:

   Monday = Date - Weekday(Date, vbMonday) + 1
   Friday = Date - Weekday(Date, vbMonday) + 5


-Dennis Borg
P.S. This is just an assist for Ross
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Expert Comment

by:dovholuk
ID: 6324471
i THOUGHT + 6 was silly! thanks for clarifying saturday is the 28th, not friday! duh for me...

as for your "more simple" code... it's the exact same thing i posted above (except of course that i should have used 5 and not 6 lol) just arranged differently...
a-b+c still equals -b+a+c still equals a+c-b etc. it just depends on how you're looking at it. ;)

dovholuk

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Expert Comment

by:DennisBorg
ID: 6324616
>i THOUGHT + 6 was silly! thanks for clarifying saturday
>is the 28th, not friday! duh for me...

That's alright. I just realized that I was thinking today was Friday and that it's only Thursday. Wishful thinking, I guess!



>as for your "more simple" code... it's the exact same
>thing i posted above (except of course that i
>should have used 5 and not 6 lol) just arranged
>differently...
>a-b+c still equals -b+a+c still equals a+c-b etc. it just
>depends on how you're looking at it. ;)

Yes, you are correct. I guess I should have said "more intuitive" than "more simple". I had to look at it a couple times to figure it out; but the second look on my part is probably largely due to moving myself, helping a friend load the truck for a move, and a couple dozen other irons in the fire. I'm looking forward to the weekend to catch up on some much-needed sleep!


-Dennis Borg
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Author Comment

by:bozo7
ID: 6325461
When I put the following in the expression to see what value it returns the query prompts me for the Date Parameter and the the vbMonday parameter.

date - weekday(now, vbMonday) + 1

What do I have wrong?

I can put this expresion in the field expresion and it should work for testing right? Even when I put in the Criteria section it prompts me for the parameters.
I will keep trying to figure it out.

Thanks,
Ross
0
 
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Author Comment

by:bozo7
ID: 6325466
I have gotten it so it only asks me for the vbMonday parameter.
Here is the expresion:
 Date()-Weekday(Now(),[vbMonday]+1)

I am using the expresion builder is that Ok?
Ross
0
 
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Author Comment

by:bozo7
ID: 6325469
expresion should read
Date()-Weekday(Now(),[vbMonday]) + 1
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Accepted Solution

by:
dovholuk earned 200 total points
ID: 6325513
vbmonday is an access constant. replace it with 2, ( 2 = vbmonday, 1 = vbsunday, etc.) the literal value for vbMonday.

dovholuk

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Author Comment

by:bozo7
ID: 6326655
Thank You,

Ross
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