Re: Fixing faulty dates - select on day part of a date field

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От Gary Stainburn
Тема Re: Fixing faulty dates - select on day part of a date field
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Msg-id E169RhA-0006sD-00@stan.ringways.co.uk
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Ответ на Re: Fixing faulty dates - select on day part of a date field  (Markus Bertheau <twanger@bluetwanger.de>)
Ответы Re: Fixing faulty dates - select on day part of a date field  (Markus Bertheau <twanger@bluetwanger.de>)
Re: Fixing faulty dates - select on day part of a date field  ("Christopher Kings-Lynne" <chriskl@familyhealth.com.au>)
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On Thursday 29 November 2001 12:06 pm, Markus Bertheau wrote:
> On Thu, 2001-11-29 at 12:31, Gary Stainburn wrote:
> > I've got a problem with dates on one of my tables. I've been inserting
> > dates in the format 'dd/mm/ccyy' which for the days 13-31 for each month
> > worked fine.
> >
> > The problem I have is that for the days 01-12 for each month, the date
> > was interpretted as 'mm/dd/ccyy'.  Now I know about it I need to fix it.
> >
> > Two questions.
> >
> > 1) how can I select on part of a date?  I need to select all records
> > where the day is not > 12.
>
> select date_part('part', attribute)
>
> where part is one of week, day, year ans so on
>
> standards compliant way is
>
> select extract(part from attribute)
>
> so:
>
> select * from table where date_part('day', attribute) < 13;
>
> > 2) can I do this in a single update, i.e. can I do something around
> >
> > update calls set xdate =  ???? where ??????;
>
> What exactly do you want to achieve?

Thanks for the information.  Basically what I want to be able to do is 
correct the dates that are wrong by swapping the day and month parts.

For example, a date entered as 23/11/2000 is correctly stored as 2000/11/23 
while a date entered as 07/11/2000 (7th Nov 2000) is incorrectly stored as 
2000/07/11.  I need to be able to access it using the select you gave above, 
and then re-store the date as 2000/11/07.

something like:

update calls set date_part('day',xdate) = date_part('month',xdate),                 date_part('month',xdate) =
date_part('day',xdate)      where date_part('day',xdate) < 13 and xdate < '2001/11/08';
 

(2001/11/08 is when I found/fixed the insert problem)

>
> Markus Bertheau

-- 
Gary Stainburn
This email does not contain private or confidential material as it
may be snooped on by interested government parties for unknown
and undisclosed purposes - Regulation of Investigatory Powers Act, 2000     


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