How to determine the type of a child object from the SQLAlchemy model?

Question

I have the following SQLAlchemy model:

class PersonConsent(ModelAbstract):
    __tablename__ = "personConsent"
    id = db.Column(db.Integer, primary_key=True)
    patientId = db.Column(db.Integer, db.ForeignKey("person.id"))
    authorizedPersonId = db.Column(db.Integer, db.ForeignKey("person.id"))
    attachmentId = db.Column(db.Integer)

    # Additional models
    authorizedPerson = db.relationship("Person", foreign_keys=[authorizedPersonId])
    consents = db.relationship("PersonConsentType", 
                secondary=personConsentToTypeMap,
                primaryjoin=id==personConsentToTypeMap.c.consentId,
                secondaryjoin=PersonConsentType.id==personConsentToTypeMap.c.consentTypeId)

Given just the PersonConsent model how do I determine the model of items that make up the consents field?

For example, something like

type(PersonConsent.consents) == PersonConsentType

Jakobovski · Accepted Answer · 2015-06-07 23:29:19Z

4

Use the inspect function and follow the correct attributes to get the target model for a relationship.

from sqlalchemy import inspect

mapper = inspect(PersonConsent)
property = mapper.relationships['consents']
target = property.mapper.class_

assert target is PersonConsentType

You can technically get to this by doing PersonConsent.consents.property.mapper.class_, but it's less general. You could use the inspection above to, for example, loop over all the relationships, even if you don't know their names.

edited Jun 7, 2015 at 23:29

Jakobovski

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answered Dec 23, 2014 at 3:21

davidism

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