当前上下文为空时如何处理sqlalchemy onupdate?

Mas*_*gol 3 python sqlalchemy

我有一个文章的模型,根据它的标题将有slug,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'])
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slug取得冠军的价值.所以,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其标题时,会引发此异常

(builtins.KeyError) 'title' [SQL: 'UPDATE article SET content=?, slug=?,
updated_at=? WHERE article = ?'] [parameters: [{'article_id': 1,
'content': 'blah blah blah'}]]
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我猜是因为current_params不包含title.如果,我改变lambda那里和使用if,slug将是None.那么我怎么能处理这个并保持slug值与它的标题相匹配?

r-m*_*m-n 5

你可以使用validates()装饰器

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
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事件

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
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