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 <mark.a.b...@gmail.com> 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 <mark....@gmail.com> 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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>>
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