Re: [GENERAL] Hers's one with the COPY command ...
От | Lincoln Yeoh |
---|---|
Тема | Re: [GENERAL] Hers's one with the COPY command ... |
Дата | |
Msg-id | 3.0.5.32.19991125091343.008d3da0@pop.mecomb.po.my обсуждение исходный текст |
Ответ на | Hers's one with the COPY command ... (The Hermit Hacker <scrappy@hub.org>) |
Список | pgsql-general |
I assume you are copying batches of data at the same time, so hardcoding in the datetime wouldn't be a big problem right? Kludgy but should work. Maybe if you did 'now' it might work. Or force it as 'now'::datetime. Definitely not an expert- plenty to learn about SQL and Postgres, but since no one has made any suggestions... Cheerio, Link. At 09:07 PM 22-11-1999 -0400, you wrote: > >I have a table created as: > >============= >DROP TABLE referer_raw_data; >CREATE TABLE referer_raw_data ( > counter_id int4, > referer_url_domain varchar(256), > referer_counter int4, > date_added datetime default now()); >============= > >I wish to copy a stream of data to it as: > >5\thttp://thissite\t3\t... > >where the date *isn't* part of the copy, since I want to set it as the >date that the record was added...is this possible? now, from the docs, I >realize that it will ignore the default, which I'm okay with, but is there >a way of passing it now(), or something similar, to set the data? > >basically, I want to stream web data to the backend as fast as the backend >will take it, with COPY TO being faster then INSERT INTO, but don't want >to have to "create a date" to insert at the same time... > >Marc G. Fournier ICQ#7615664 IRC Nick: Scrappy >Systems Administrator @ hub.org >primary: scrappy@hub.org secondary: scrappy@{freebsd|postgresql}.org > > >************ > > >
В списке pgsql-general по дате отправления: