How to avoid inserting duplicate entries when adding values via a sqlalchemy relationship?

Question

Let's assume we have two tables in a many to many relationship as shown below:

class User(db.Model):
  __tablename__ = 'user'
  uid = db.Column(db.String(80), primary_key=True)
  languages = db.relationship('Language', lazy='dynamic',
                              secondary='user_language')

class UserLanguage(db.Model):
  __tablename__ = 'user_language'
  __tableargs__ = (db.UniqueConstraint('uid', 'lid', name='user_language_ff'),)

  id = db.Column(db.Integer, primary_key=True)
  uid = db.Column(db.String(80), db.ForeignKey('user.uid'))
  lid = db.Column(db.String(80), db.ForeignKey('language.lid'))

class Language(db.Model):
  lid = db.Column(db.String(80), primary_key=True)
  language_name = db.Column(db.String(30))

Now in the python shell:

In [4]: user = User.query.all()[0]

In [11]: user.languages = [Language('1', 'English')]

In [12]: db.session.commit()

In [13]: user2 = User.query.all()[1]

In [14]: user2.languages = [Language('1', 'English')]

In [15]: db.session.commit()

IntegrityError: (IntegrityError) column lid is not unique u'INSERT INTO language (lid, language_name) VALUES (?, ?)' ('1', 'English')

How can I let the relationship know that it should ignore duplicates and not break the unique constraint for the Language table? Of course, I could insert each language separately and check if the entry already exists in the table beforehand, but then much of the benefit offered by sqlalchemy relationships is gone.

Answer 1

The SQLAlchemy wiki has a collection of examples, one of which is how you might check uniqueness of instances.

The examples are a bit convoluted though. Basically, create a classmethod get_unique as an alternate constructor, which will first check a session cache, then try a query for existing instances, then finally create a new instance. Then call Language.get_unique(id, name) instead of Language(id, name).

I've written a more detailed answer in response to OP's bounty on another question.