Thanks! That is the what i want.
On 6月22日, 午後11:40, "Michael Bayer" <mike...@zzzcomputing.com> wrote:
> Alisue wrote:
>
> > Oops. I found the solution with my self.
>
> > like below.
>
> > c.paginator = paginate.Page(
> > meta.Session.query(Article).select_from(query.subquery()),
> > page=int(request.params.get('page',1)),
> > items_per_page=int(request.params.get('items_per_page', 10))
> > )
>
> > But why can't I do like below? or similar way?
>
> > query.select_from(query.subquery())
>
> that is what query.from_self() does.
>
>
>
> > On 6月22日, 午後6:34, Alisue
> > <hello.goodbye.by.beat...@gmail.com> wrote:
> >> I have a Article table may relate with a Room table and I want to find
> >> Article sometime with Room.name.
> >> So I wrote the code like below.
>
> >> query = meta.Session.query(Article)
> >> query = query.outerjoin('rooms')
> >> if room_name is not None:
> >> query = query.filter(Room.name == room_name)
> >> query = query.group_by(Article.id)
> >> return query
>
> >> A result is correct. I got all Article name without room_name and
> >> Articles which has rooms named as room_name.
>
> >> However, when I use this query with webhelpers.paginate(http://
> >> beta.pylonshq.com/docs/ja/0.9.7/thirdparty/webhelpers/paginate/
> >> #webhelpers.paginate.Page), everything goes wrong. 'paginator' doesn't
> >> work correct.
>
> >> So I chacked logs and find this output.
> >> 'SELECT COUNT(1) AS count_1 FROM articles LEFT OUTER JOIN rooms ON
> >> articles.id = rooms.article_id GROUP BY articles.id'
>
> >> This is might made by .count() function of sqlalchemy in somewhere in
> >> webhelpers.paginate code I think. However what I want to get is
> >> 'SELECT COUNT(1) AS count_1 FROM (SELECT * FROM articles LEFT OUTER
> >> JOIN rooms ON articles.id = rooms.article_id GROUP BY articles.id)'
>
> >> I have no idea to fix this problem. Anyone? thank you.
>
> >> SQLAlchemy: 0.5.4
> >> Pylons: 0.9.7
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