That's perfect. Thanks!
On Thursday, 10 April 2014 11:38:24 UTC+1, Simon King wrote:
>
> You can also use sqlalchemy.literal, which returns an object that you
> can treat like a column:
>
> sqlalchemy.literal('string').like('whatever')
>
> You may also be interested in the 'startswith' shortcut, which calls
> .like under the hood
>
> sqlalchemy.literal('string').startswith(yourcolumn)
>
> Simon
>
> On Thu, Apr 10, 2014 at 11:35 AM, Mark Bird <[email protected]<javascript:>>
> wrote:
> > I'm sorry, forget my last response, I see what you mean now.
> >
> > Thanks a lot!
> >
> >
> > On Thursday, 10 April 2014 11:31:23 UTC+1, Gunnlaugur Briem wrote:
> >>
> >> Hi,
> >>
> >> See ColumnElement docs:
> >>
> >>
> >>
> http://docs.sqlalchemy.org/en/rel_0_9/core/sqlelement.html#sqlalchemy.sql.expression.ColumnElement
>
> >>
> >> ... for your specific example you can call .like(...) on column
> clauses:
> >>
> >> >>> print Column('foo', Text).like('bar%baz')
> >> foo LIKE :foo_1
> >>
> >> More generally, if you wanted some operator other than LIKE, existing
> in
> >> your DB dialect but not yet in SQLAlchemy, then you can use .op(...):
> >>
> >> >>> print Column('foo', Text).op('FNORD')('foo%')
> >> foo FNORD :foo_1
> >>
> >> Cheers,
> >>
> >> Gulli
> >>
> >>
> >>
> >> On Thu, Apr 10, 2014 at 12:19 PM, Mark Bird <[email protected]>
> wrote:
> >>>
> >>> I can't seem to find a way to do this without passing raw SQL to
> >>> .filter()
> >>>
> >>> I could just do:
> >>>
> >>> .filter(column == func.substring('string', 1,
> func.char_length(column)))
> >>>
> >>> but is it possible to do it with LIKE?
> >>>
> >>> I.e. I need to return all rows that match the beginning of a string,
> so
> >>> for 'string' I could match 's', 'st', 'str', etc.
> >>>
> >>> Thanks,
> >>>
> >>> Mark.
> >>>
> >>>
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