当前上下文为空时如何处理sqlalchemy onupdate?
问题内容:
我有一个文章模型,该模型将基于其标题插入,该模型如下所示:
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
正在获取标题的价值。因此,每次文章slug
都会与其标题匹配。但是,当我在不更改标题的情况下编辑内容时,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'}]]
我想是因为其中current_params
不包含title
。如果我更改了lambda
那里并使用if
,则将是None
。那么我该如何处理并保持slug值与其标题匹配?
问题答案:
您可以使用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
或事件:
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