Re: [office-formula] Who's doing what this week,Redux (including a request to Stefan Nikolas)

[Eike Rathke <erack@sun.com>]

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.