Re: Week of the Year?
От | John McKown |
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Тема | Re: Week of the Year? |
Дата | |
Msg-id | Pine.LNX.4.21.0008112119390.15236-100000@linux2.johnmckown.net обсуждение исходный текст |
Ответ на | Week of the Year? ("David Lloyd-Jones" <david.lloyd-jones@attcanada.ca>) |
Список | pgsql-sql |
Try using the function date_part such as: select date_part('week',now()); "and the days that are in that week" I guess want to answer a question such as: Given a date, what is first date in that same week, and what is the last date in that week. There are a couple of approaches to this. My first was: select to_date(date_part('year',now()),'YYYY')+(7*date_part('week',now())); and the above +6 to the the last day of the week. Another approach for this same question is much simplier (if the question is indeed what you are asking) select now()-date_part('dow',now()); This last select gives the Sunday for the current week. To get the Saturday, simply: select now()-date_part('dow',now())+6; Of course, replace the now() with whatever contains the date or timestamp. John McKown > I'm probably staring right at it. (One of the difficulties with RTFMing, is > having too many docs!) > > Is there anything in the API that produces the week of the year, from 1 to > 52 or 53 depending on the week of the year, and the days that are in that > week? > > Many thanks. > > -dlj. > > >
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