2015-02-09 135 views
0

所以,如果我有两个SQLAlchemy的表像如何使用子查询筛选一对多关系中的sqlalchemy查询?

class Parent 
    id = Column(Integer, primary_key=True) 
    children = relationship('Child', lazy='joined', backref=backref('parent', lazy='joined')) 
class Child 
    id = Column(Integer, primary_key=True) 
    age = Column(Integer) 

我如何找到所有已至少有一个孩子随着年龄> 10的父母?

我已经试过这样的事情,虽然这不工作:

Session.query(Parent.id).filter(func.count(Parent.children.filter(Child.age >= 10)) > 0) 
+0

尝试'backref = backref('parent',lazy ='dynamic'))' – 2015-02-09 21:55:01

回答

2

假设这种模式:

class Parent(Base): 
    id = Column(Integer, primary_key=True) 
    name = Column(String) 
    children = relationship('Child', lazy='joined', 
          backref=backref('parent', lazy='joined')) 

class Child(Base): 
    id = Column(Integer, primary_key=True) 
    parent_id = Column(ForeignKey(Parent.id)) 
    age = Column(Integer) 

使用any结构得到过滤工作:

q = (session.query(Parent) 
    .filter(Parent.children.any(Child.age >= 10)) 
    ) 

for p in q: 
    print("{}".format(p)) 
    for c in p.children: 
     print(" {}".format(c))