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Say, we have the following relationships: a person can have many email addresses a email service provider can (obviously) serve multiple email address So, it's a many to many relationship. I have three tables: emails, providers, and users. Emails have two foreign ids for provider and user. Now, given a specific person, I want to print all the email providers and the email address it hosts for this person, if it exists. (If the person do not have an email at Gmail, I still want Gmail be in the result. I believe otherwise I only need a left inner join to solve this.) I figured out how to do this with the following subqueries (following the sqlalchemy tutorial): email_subq = db.session.query(Emails).\ filter(Emails.user_id==current_user.id).\ subquery() provider_and_email = db.session.query(Provider, email_subq).\ outerjoin(email_subq, Provider.emails).\ all() This works okay (it returns a 4-tuple of (Provider, user_id, provider_id, email_address) , all the information that I want), but I later found out this is not using the Flask BaseQuery class, so that pagination provided by Flask-SQLAlchemy does not work. Apparently db.session.query() is not the Flask-SQLAlchemy Query instance. I tried to do Emails.query.outerjoin[...] but that returns only columns in the email table though I want both the provider info and the emails. My question : how can I do the same thing with Flask-SQLAlchemy so that I do not have to re-implement pagination that is already there? I guess the simplest option at this point is to implement my own paginate function, but I'd love to know if there is another proper way of doing this.

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