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Subject: Re: [office-formula] WEEKNUM reference
Wheeler: > > 1. Simple/absolute weeks > > We can cover the "absolute/simple" week number system with a trivial modification to WEEKNUM. Here's a proposal: if WEEKNUM's second parameter is "0", then the simple week number is returned. That would mean that WEEKNUM(d;0) == WEEKNUM(x; WEEKDAY(DATE(1;1;YEAR(x))). Eike: > Apart from that it would be DATE(YEAR(x);1;1) instead, I don't see why > WEEKNUM(d;0) would be equal to WEEKNUM(x;n) with n!=0. Ooops, I changed the unbound variable midways through. Bad me! Let me try again; what I meant was: WEEKNUM(x;0) == WEEKNUM(x; WEEKDAY(DATE(1;1;YEAR(x))) Eike: > Actually quite a few locales not using ISO 8601 week numbering use > minimum-number-of-days 1, resulting in the first week of the year being > the week where the first start-of-week-day of that year is, which would > be covered by your 1..7 proposal above. > > Some, similar to ISO 8601, use minimum-number-of-days 4, but with Sunday > as start-of-week-day. These would need extra handling. Currently the > locales are *-CA Canada and *-MT Malta, according to the CLDR > supplementalData.xml Eek. Okay, I stand corrected. Okay, I think there's wide agreement that we need to _AT LEAST_ add support for 1..7 as the second parameter. Should we add the optional 3rd parameter for "minimum number of days in a week?? Define "0" for simple years? --- David A. Wheeler
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