我有以下两个类:
class Product(db.Model):
__tablename__ = 'product'
ProductID = db.Column(db.Integer, primary_key=True)
StartOperatorID = db.Column(db.Integer, db.ForeignKey('user.UserID'), nullable=False)
StartOperator = db.relationship("User", foreign_keys=[StartOperatorID])
WorkBenchID = db.Column(db.Integer, db.ForeignKey('workbench.WorkBenchID'))
WorkBench = db.relationship(WorkBench)
class WorkBench(db.Model):
__tablename__ = 'workbench'
WorkBenchID = db.Column(db.Integer, primary_key=True)
但是,我得到以下NameError,因为WorkBench定义如下:Product:
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-1-3a482de7fdcf> in <module>()
----> 1 from example import models
/home/Sin5k4/example/models.py in <module>()
16
---> 17 class Product(db.Model):
18 ProductID = db.Column(db.Integer, primary_key=True)
/home/Sin5k4/example/models.py in Product()
21 WorkBenchID = db.Column(db.Integer, db.ForeignKey('workbench.WorkBenchID'))
---> 22 WorkBench = db.relationship(WorkBench)
23
NameError: name 'WorkBench' is not defined
来自.NET背景,我真的很挣扎。定义与此类的关系的正确方法是什么?
答案 0 :(得分:5)
当解释Python程序时,类定义按顺序执行,包括类级变量的定义(例如该关系)。为了允许关系的无序定义,除了直接传递类之外,SQLAlchemy还有一些其他的语义。实际上,您的其他一些关系已经在使用此表单:传递字符串名称而不是类对象。您还可以传递一个评估为类对象的lambda,或者修复它。
# by class name (preferred in most cases)
WorkBench = db.relationship('WorkBench')
# or by lambda
WorkBench = db.relationship(lambda: WorkBench)