Hi Robert,
On Wednesday, 2007-02-14 18:54:45 -0500, Robert Weir wrote:
> I'm thinking of the default case, with no holidays and no user-defined
> workweek.
>
> In Excel today, NETWORKDAYS(TODAY(),NOW()) is always going to be 1.0. If
> we allow fractional days in the calculation, then this function will give
> a different answer depending on the time of day, right?
Ah, no, sorry, seems I wasn't clear on that. I don't want to allow
fractional start and end dates, but if a matrix passed as workdays
allowed fractional values [0,1] the function would calculate
a fractional for these days of the week. For example,
=NETWORKDAYS(TODAY();TODAY()+7;0;{1;1;1;1;1;0.5;0}) would give 5.5.
Analogous for the WORKDAY function.
Eike
--
Automatic string conversions considered dangerous. They are the GOTO statements
of spreadsheets. --Robert Weir on the OpenDocument formula subcommittee's list.
