On Jun 6, 2013, at 12:56 PM, Ladislav Lenart <lenart...@volny.cz> wrote:
> Hello. > > I have already solved the issue by using subquery: > > SELECT > t.id AS t_id, > t.rownum AS t_rownum > FROM ( > SELECT > FROM > foo.id AS id, > row_number() OVER (ORDER BY foo.id) AS rownum > ) AS t > WHERE rownum % 50 = 1 > > I have just tried your suggestion about using HAVING instead of WHERE, but > that > fails with the same error. Thus a label cannot be used inside a query. > > However, I am still curious whether the original WindowedRangeQuery recipe at > > http://www.sqlalchemy.org/trac/wiki/UsageRecipes/WindowedRangeQuery > > works or also has this error. the recipe as you noted uses from_self(), which means, "wrap myself in a subquery", so that's where the necessary subquery is applied. -- You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy?hl=en. For more options, visit https://groups.google.com/groups/opt_out.
