Re: Order by YYYY MM DD in reverse chrono order trouble

Hello,

I'd love to be able to do that, but I cannot just ORDER BY uu.add_date,
because I do not have uu.add_date in the SELECT part of the statement.
The reason I don't have it there is because I need distinct YYYY MM DD
values back.
Is there a trick that I could use to make this more elegant?

Thanks,
Otis

--- Edmund Bacon <ebacon(at)onesystem(dot)com> wrote:
> Is there some reason you can't do this:
>
> SELECT DISTINCT
> date_part('year', uu.add_date), date_part('month', uu.add_date),
> date_part('day', uu.add_date)
>
> FROM uus INNER JOIN ui ON uus.user_id=ui.id INNER JOIN uu ON
> ui.id=uu.user_id
> WHERE uus.x_id=1
>
> ORDER BY
> uu.add_date DESC;
>
> This might be faster, as you only have to sort on one field, and I
> think it should give the desired results
>
> ogjunk-pgjedan(at)yahoo(dot)com wrote:
>
> >Hello,
> >
> >I am trying to select distinct dates and order them in the reverse
> >chronological order. Although the column type is TIMESTAMP, in this
> >case I want only YYYY, MM, and DD back.
> >
> >I am using the following query, but it's not returning dates back in
> >the reverse chronological order:
> >
> >SELECT DISTINCT
> > date_part('year', uu.add_date), date_part('month', uu.add_date),
> > date_part('day', uu.add_date)
> >
> >FROM uus INNER JOIN ui ON uus.user_id=ui.id INNER JOIN uu ON
> >ui.id=uu.user_id
> >WHERE uus.x_id=1
> >
> >ORDER BY
> >date_part('year', uu.add_date), date_part('month', uu.add_date),
> >date_part('day', uu.add_date) DESC;
> >
> >
> >This is what the above query returns:
> >
> > date_part | date_part | date_part
> >-----------+-----------+-----------
> > 2004 | 2 | 6
> > 2004 | 4 | 20
> >(2 rows)
> >
> >
> >I am trying to get back something like this:
> >2004 4 20
> >2004 4 19
> >2004 2 6
> >...
> >
> >My query is obviously wrong, but I can't see the mistake. I was
> >wondering if anyone else can see it. Just changing DESC to ASC, did
> >not work.
> >
> >Thank you!
> >Otis
> >
> >
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>
> --
> Edmund Bacon <ebacon(at)onesystem(dot)com>
>