Question

I'm trying to add a "slug" field to my db.Model class Post, in which it takes the title of the post and returns it in a better format for a URL.

class Post(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    title = db.Column(db.String(64), index=True, unique=True)
    body = db.Column(db.Text)
    timestamp = db.Column(db.DateTime, index=True, default=datetime.utcnow)
    user_id = db.Column(db.Integer, db.ForeignKey('user.id'))

    def __repr__(self):
        return '<Post {}>'.format(self.title)

    @hybrid_property
    def slug(self):
        result = []
        punct_re = re.compile(r'[\t !"#$%&\'()*\-/<=>?@\[\\\]^_`{|},.]+')
        for word in punct_re.split(self.title.lower()):
            if word:
                result.append(word)
        return '-'.join(result)

I can reference the slug by Post.slug, but when I try to query it, as Post.query.filter_by(slug='test-slug-1'), I get the message

sqlalchemy.exc.InvalidRequestError: Entity <class 'app.models.Post'> has no property 'slug'

From what I understand I need a hybrid expression, but the examples I've seen have made it more confusing. Any ideas on how to filter by the hybrid property slug, or am I going about generating this slug incorrectly?

Python SQLAlchemy Hybrid Property and Expression

0 个答案: