Question

我有一个基于它的标题的文章模型，模型是这样的：

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, Integer, String, Text

Base = declarative_base()


class Article(Base):

    __tablename__ = 'article'

    id = Column(Integer, primary_key=True)
    title = Column(String(100), nullable=False)
    content = Column(Text)
    slug = Column(String(100), nullable=False,
                  default=lambda c: c.current_params['title'],
                  onupdate=lambda c: c.current_params['title'])

slug正在获得头衔的价值。因此，每次文章slug都会匹配它的标题。但是，当我编辑内容而不更改它的标题时，这个异常被提出

(builtins.KeyError) 'title' [SQL: 'UPDATE article SET content=?, slug=?,
updated_at=? WHERE article = ?'] [parameters: [{'article_id': 1,
'content': 'blah blah blah'}]]

我猜是因为current_params并不包含title。如果，我改变了那里lambda并使用if，slug将是None。那我该怎么办？这和slug值保持匹配吗？

Answer 1

您可以使用validates() decorator

from sqlalchemy.orm import validates

class Article(db.Model):
    __tablename__ = 'article'
    id = db.Column(db.Integer, primary_key=True)
    title = db.Column(db.String(100), nullable=False)
    content = db.Column(db.String)
    slug = db.Column(db.String(100), nullable=False)

    @validates('title')
    def update_slug(self, key, title):
        self.slug = title
        return title

或events

from sqlalchemy import event

class Article(db.Model):
    __tablename__ = 'article'
    id = db.Column(db.Integer, primary_key=True)
    title = db.Column(db.String(100), nullable=False)
    content = db.Column(db.String)
    slug = db.Column(db.String(100), nullable=False)

@event.listens_for(Article.title, 'set')
def update_slug(target, value, oldvalue, initiator):
    target.slug = value

当前上下文为空时如何处理sqlalchemy onupdate？

1 个答案: