how to force a filter() to match exact cases (case sensitivity) in sqlalchemy?

Question

I am using sqlalchemy with mysql database. When I am using following query on User object :

session.query(User).filter(User.name == 'admin').all()

I get all the results which have usename as 'Admin', 'admin', 'ADMIN' (basically all case-insensitive 'admin'). I would like to know how can I force a filter() to filter exact matches (without ignoring cases)?

Update : Actually I just came to know that mysql does not allow case-sensitive columns for varchar() datatype. So the simplest solution would be to force a column to be case-sensitive while declaring in mysql, as :

  `name` VARCHAR(255) BINARY NULL UNIQUE,

But I would still love to know how can I force a filter to match the results exactly (not ignoring cases). Is there anyway using some inbuilt (or custom, if possible) sqlalchemy func?

Answer 1

select * from users where binary name = 'AdMin';
select * from users where binary name = 'admin';


from sqlalchemy import func
User.query.filter(User.name == func.binary('AdMin')).count()
User.query.filter(User.name == func.binary('admin')).count()

Answer 2

Generally, what SQLAlchemy does is construct a SQL string query out of the expressions. Typing freely your above expression would yield something like this:

SELECT user.* FROM user WHERE user.name = 'admin'

Of course, user.* would replaced by an explicity enumeration of all user columns.

The thing is that it depends what this query returns: If it compares case-insensitively, there is not much SQLAlchemy can do. From what I gather looking around, you can explicitly construct a query using WHERE BINARY user.name = 'admin'. However, I am not aware of a SQLAlchemy equivalent, so you might be stuck with building a literal string here.

Anyways, I would suggest that you ask on the SQLAlchemy list for the particular solution (especially, I think it would be interesting to know a way of intergrating the BINARY into a WHERE query).