CODEWITHSUNDEEP
pythonsqlalchemy
mas
mas

Reputation: 1245

Order by count on many to many relationship without returning a tuple with count

With the following simplified classes and tables

tags = db.Table(
        "tags",
        db.Column("tag_id", db.Integer, db.ForeignKey("tag.id")),
        db.Column("post_version_id", db.Integer, db.ForeignKey("post_version.id"))
        )

class Tag(db.Model):
    id = db.Column(db.Integer, primary_key=True)

class PostVersion(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    current = db.Column(db.Boolean, default=False, index=True)

I am trying to issue a call for all Tag's ordered by the number of PostVersion's that are both connected via tags and possess the value current=True.

I have developed the following query:

db.session.query(Tag,
        db.func.count(PostVersion.id).label("total")
        ).outerjoin(tags
        ).outerjoin(PostVersion,    
                and_(PostVersion.id==tags.c.post_version_id,
                PostVersion.current==True)
        ).group(Tag).order_by("total DESC").all()

Which yields the following (correct) results:

[(<Tag 8: original>, 136),
 (<Tag 16: constance-garnett>, 136),
 (<Tag 3: explanation>, 3),
 (<Tag 2: definition>, 1),
 (<Tag 14: translation>, 1),
 (<Tag 1: biblical>, 0),
 (<Tag 4: homage>, 0),
 (<Tag 5: intertextuality>, 0),
 (<Tag 6: meter>, 0),
 (<Tag 7: mythology>, 0),
 (<Tag 9: political>, 0),
 (<Tag 10: cultural>, 0),
 (<Tag 11: reference>, 0),
 (<Tag 12: shakespeare>, 0),
 (<Tag 13: technical-issues>, 0),
 (<Tag 15: context>, 0)]

Except that I have to do additional modifications to the results in order to suppress the tuple-based output which includes the results of db.func.count(PostVersion.id).label("total") and I am entirely unsure of how to rewrite the query to suppress just that.

How can I formulate my query to just return the objects in the order this query returns?

Upvotes: 0

Views: 32

Answers (1)

Ilja Everil&#228;
Ilja Everil&#228;

Reputation: 52929

The solution is simple: move the count expression to the ORDER BY clause:

db.session.query(Tag).\
    outerjoin(tags).\
    outerjoin(PostVersion,
              and_(PostVersion.id==tags.c.post_version_id,
                   PostVersion.current==True)).\
    group_by(Tag.id).\
    order_by(db.func.count(PostVersion.id).desc()).\
    all()

Upvotes: 1

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