CODEWITHSUNDEEP
pythonsqlalchemy
Friendly King
Friendly King

Reputation: 2476

SQLAlchemy filter() return value

I'm having one of those days.

I want to check if a particular row exists in my DB, and if it does, return True. If not, return False.

A brief example:

contact_object = Contact.query.filter(Contact.user_id == user_id and Contact.contact_id == contact_id).first()

This should return exactly 1 row (which it does). Here's what I would like:

if contact_object:
    print "YEY, it was found"
else:
    print "Nope, not found"

However, contact_object always returns True, and I always get a "YEY, it was found"

(Note that user_id and contact_id are variables that I have defined previously.)

How do I check for a row not found versus a row found? I read the relevant documentation to no avail...but I feel like I'm surely missing something simple?

Thanks for your help.

Upvotes: 2

Views: 6290

Answers (1)

David Maust
David Maust

Reputation: 8270

I don't believe you can use and in a filter expression in SQL Alchemy that way:

http://docs.sqlalchemy.org/en/rel_0_8/orm/tutorial.html#common-filter-operators

contact_object = Contact.query \
    .filter(Contact.user_id == user_id) \
    .filter(Contact.contact_id == contact_id) \
    .first()

Upvotes: 3

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