Python 在to_json方法中加载关系
我有一个相当基本的Python 在to_json方法中加载关系,python,python-3.x,flask,sqlalchemy,flask-sqlalchemy,Python,Python 3.x,Flask,Sqlalchemy,Flask Sqlalchemy,我有一个相当基本的CRUDMixin class CRUDMixin(object): """ create, read, update and delete methods for SQLAlchemy """ id = db.Column(db.Integer, primary_key=True) @property def columns(self): return [ c.name for c in self.__table__.colu
CRUDMixin
class CRUDMixin(object):
""" create, read, update and delete methods for SQLAlchemy """
id = db.Column(db.Integer, primary_key=True)
@property
def columns(self):
return [ c.name for c in self.__table__.columns ]
def read(self):
""" return json of this current model """
return dict([ (c, getattr(self, c)) for c in self.columns ])
# ...
对于像Article
类这样的将子类化的类,它可能与另一个类有关系,如下所示:
author_id = db.Column(db.Integer, db.ForeignKey('users.id'))
唯一真正的问题是,它不会在json中返回任何用户详细信息。理想情况下,json应该如下所示:
{
'id': 1234,
'title': 'this is an article',
'body': 'Many words go here. Many shall be unread. Roman Proverb.',
'author': {
'id': 14
'name': 'Thor',
'joined': October 1st, 1994
}
}
现在,它只会给出作者id:14
我是否可以检测列是否是关系并以这种方式将其作为json加载?您必须通过添加以下内容来设置整个关系
author = db.relationship("Author") # I assume that you have an Author model
然后,对于json结果,您有不同的处理关系的方法
请看这两个回答:
您还可以看看flask restful,它提供了一个方法/装饰器(marshal_with),可以用嵌套对象(关系)很好地封送结果