Public Function paydays(lyear As Long, _
lmonth As Long, dtfirstpayday, lpayperiod)
Dim dtStart As Date, dtEnd As Date
Dim dtStart1 As Date, dtEnd1 As Date
dtStart1 = dtfirstpayday
dtEnd1 = DateSerial(lyear, lmonth, 0)
dtEnd = DateSerial(lyear, lmonth + 1, 0)
lNumdays = dtEnd1 - dtStart1
lpays = lNumdays \ lpayperiod
dtStart = dtStart1 + lpayperiod * lpays
lNumdays = dtEnd - dtStart
paydays = lNumdays \ lpayperiod
End Function
seems to work.
--
Regards,
Tom Ogilvy
"dbmathis" wrote in
message ...
I need User defined function to count pay days in current month. Let's
see if I can explain this idea right the first time. Big grin
I am in need of a user defined function like the following:
-paydays(year, month, first pay, pay period)
paydays() output = # of paydays in current month based on:
-
year = -current year- format = YYYY
month = -current month- format = MM
first pay = -date of first pay check- format = MM/DD/YYYY
pay period = -days between pay days- format = DD
assumptions are that the person gets paid at least once a month. first
pay can be any date <= current month and current year. current month
and current year can be any date = first pay.
Example:
paydays(2005, 10, "8/29/05", 14)
would spit out a 2
-------------------------
I am creating a cashflow for my personal finances. I need a user
defined function that based on the above args can dynamically determine
how many paydays are in the current month.
I get paid every 7 days and my wife gets paid every 14 days. This means
I get paid 4 or 5 times a month and that my wife gets paid 2 or 3 time a
month. I was thinking in my sleep last night about how wonderful it
would be If I or someone looking for something similiar could just
dynamically define pay period and first pay day in a function and
automatically get the number of paydays in a current month..
If any of you are up for the challenge of coding this I would greatly
appreciate it. Thank you all for being such a helpful community.
I hope I was clear and very descriptive of what my idea is. I have
trouble with this sometimes.
thanks.
--
dbmathis
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