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calculate number of days formula
Hello Rick,
Agreed and changed, thanks.
Let us find another "challenge" :-)
Regards,
Bernd
On 25 Apr., 22:10, "Rick Rothstein"
wrote:
A couple of comments about the text description for your formula solution
and for the functions I posted to you.
In the description for your formula, you allude to it being equivalent to
NETWORKDAYS, but you forgot to mention that it does not factor in holidays.
You also mentioned NETWORKDAYS is from the "analysis add-in"... actually, to
be technically correct, it is the "Analysis ToolPak" add-in. I don't know
whether you might want to mention that NETWORKDAYS is built into Excel
2007/2010 and is part of the "Analysis ToolPak" add-in for Excel 2003 and
earlier.
With regard to my functions, the first one (without the NWD parameter) also
has the requirement that D1 needs to be the earlier date and D2 the later
date. So you may want to use this version of that function instead...
Function CWD(StartDate As Date, EndDate As Date) As Long
* CWD = DateDiff("d", StartDate, EndDate) - DateDiff("ww", StartDate,
EndDate) * 2 - _
* * * * (Weekday(EndDate) < 7) + (Weekday(StartDate) = 1) +
(Weekday(StartDate, 2) < 6)
End Function
As for the presentation of both of my functions, you show them one following
the other; however, since they both have the same name, there might be a
chance that less experienced VB coders will be confused. Perhaps you could
change the lead in description to my function to this...
"Rick Rothstein came up with an elegant equivalent VBA version:"
where I left the word "elegant" in because you used it, not because I would
necessarily have described it that way. Then, you could use something like
the following text to lead into the second function...
"Rick also provide this more general version which, if the optional NWD
parameter is omitted or set to False, will return the same values as the
above version of CWD and if the optional NWD parameter is set to True, then
it will return the same results as Excel's NETWORKDAYS function (again,
without consideration for holidays):"
--
Rick (MVP - Excel)
"Bernd P" wrote in message
...
Hello Rick,
Great. Tested and published as well. No need anymore for my cwd
version. Thanks.
Regards,
Bernd
On 25 Apr., 21:15, "Rick Rothstein"
wrote:
Sorry, it looks like I implemented the NWD parameter backwards. This
should
produce the correct results for the NWD parameter equaling True and
False...
Function CWD(D1 As Date, D2 As Date, Optional NWD As Boolean) As Long
* CWD = DateDiff("d", D1, D2) - DateDiff("ww", D1, D2) * 2 - (Weekday(D2)
_
* * * * * * *< 7) + (Weekday(D1) = 1) - (Not NWD) * (Weekday(D1, 2) < 6)
End Function
And here it is with the renamed parameters to enforce the requirement
that
D1 must be the earlier date and D2 the later one...
Function CWD(StartDate As Date, EndDate As Date, _
* * * * * * *Optional NWD As Boolean) As Long
* CWD = DateDiff("d", StartDate, EndDate) - DateDiff("ww", StartDate, _
* * * * EndDate) * 2 - (Weekday(EndDate) < 7) + (Weekday(StartDate) = 1)
_
* * * * - (Not NWD) * (Weekday(StartDate, 2) < 6)
End Function
--
Rick (MVP - Excel)
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