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The last saturday of every month for any give month

Posted on 2011-02-20
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Last Modified: 2012-05-11
Thanks,
JP
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Question by:easycapital
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Expert Comment

by:dev00790
ID: 34937339
How do you want this?

E.g. when a user puts a date into a certain cell on a worksheet, to output the date of the last saturday of that month into an adjacent cell?

So for 01/02/2011 (UK date format) in cell A1 - to output  26/02/2011 in cell B1?

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Assisted Solution

by:Peter Kwan
Peter Kwan earned 100 total points
ID: 34937366
Assume your date value is at cell A1, then at other cell, use this formula:

=DATE(YEAR(A1),MONTH(A1)+1,0)-(MAX(0,WEEKDAY(DATE(YEAR(A1),MONTH(A1)+1,0),2)+1))
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Expert Comment

by:Peter Kwan
ID: 34937388
Sorry, should be

=DATE(YEAR(A1),MONTH(A1)+1,0)-MOD(MAX(0,WEEKDAY(DATE(YEAR(A1),MONTH(A1)+1,0),2)+1),7)
Book1.xlsx
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LVL 45

Expert Comment

by:patrickab
ID: 34937406
Juan,

Try:

=EOMONTH(TODAY(),1)-(8-WEEKDAY(EOMONTH(TODAY(),2)))

You will need to install the Analysis ToolPak to get EOMONTH()

Hope it does what you want.

Patrick
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Expert Comment

by:patrickab
ID: 34937411
Juan,

ps. That formula give the date for the last Saturday in March 2011. Thus the use of EOMONTH((Today(),1) as it's next month. For April 2011 you would need to use EOMONTH((Today(),2)

Patrick
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LVL 50

Accepted Solution

by:
barry houdini earned 400 total points
ID: 34937669
Hello Patrick,

That formula gives me a Wednesday, 30th March 2011

Juan,

If you want the last Saturday of the current month you could do that with this formula

=DATE(YEAR(A1),MONTH(A1)+1,1)-WEEKDAY(DATE(YEAR(A1),MONTH(A1)+1,1))

assuming today's date in A1

....or shorter with EOMONTH

=EOMONTH(A1,0)+1-WEEKDAY(EOMONTH(A1,0)+1)

If you then want a list of "last Saturdays" then assuming you have one of the above formulas in A3 then this formula in A4 copied doqwn will give you as many subsequent last Saturdays as you need

=A3+35-(DAY(A3+35)<8)*7

See attached,

To test you can try putting different dates in A1 to see what results you get.....

regards, barry
26834429.xls
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Author Closing Comment

by:easycapital
ID: 34938530
Patrick,
Good to see you again in the questions that I am posting.  On this one, I would have to agree with Barry.
Pkwan,
Thank you for the formula.
Barry,
Many thanks for the 2 approaches - extremely helpful and very well documented.

Thanks,
JP
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