String literal as the left hand side argument of 'like' operator using SQLAlchemy

Question

Im trying to construct a query using SQLAlchemy which produces something like (in oracle):

select * from users u where 'john' like u.name || '%'

to get names like 'j', 'jo', 'joh' etc.

I suppose there'a something like:

session.query(Users).filter(XXX('john').like(Users.name + '%')).all()

What should I replace XXX with?

Audrius Kažukauskas · Accepted Answer · 2013-02-25 19:55:19Z

3

Use literal construct:

session.query(Users).filter(literal('john').startswith(Users.name)).all()

answered Feb 25, 2013 at 19:55

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Comments

reptilicus · Accepted Answer · 2013-02-25 18:38:35Z

-1

I think its something like this?

Users.query.filter(Users.name.like("%john%")).all()

answered Feb 25, 2013 at 18:38

reptilicus

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