我在我的Flask應用程序中使用SQLAlchemy(0.9.4)。在應用程序中有兩個具有軟刪除支持的表。SQLAlchemy - 在自定義主連接關係的查詢中使用'aliased'
class A(SoftDeleteMixin, db.Model):
id = db.Column(db.BigInteger, primary_key=True)
b_id = db.Column(db.BigInteger, db.ForeignKey('b.id'), nullable=False)
b = soft_delete_relationship('B.id', 'A.b_id')
class B(SoftDeleteMixin, db.Model):
id = db.Column(db.BigInteger, primary_key=True)
parent_id = db.Column(db.BigInteger, db.ForeignKey('b.id'))
parent = soft_delete_relationship(remote(id), parent_id, 'B.id', 'B.parent_id')
children = soft_delete_relationship(remote(parent_id), id, 'B.parent_id', 'B.id')
SoftDeleteMixin基於LimitingQuery(https://bitbucket.org/zzzeek/sqlalchemy/wiki/UsageRecipes/PreFilteredQuery)
from sqlalchemy.orm.query import Query
class NonDeletedQuery(Query):
def get(self, ident):
return Query.get(self.populate_existing(), ident)
def __iter__(self):
return Query.__iter__(self.private())
def from_self(self, *ent):
return Query.from_self(self.private(), *ent)
def private(self):
mzero = self._mapper_zero()
if mzero is not None and hasattr(mzero, 'class_'):
soft_deleted = getattr(mzero.class_, 'soft_deleted', None)
return self.enable_assertions(False).filter(soft_deleted.is_(False)) if soft_deleted else self
else:
return self
而且soft_delete_relationship定製primaryjoin結構關係(加入非soft_deleted)。
b_parent = aliased(B)
A.query.join(A.b).outerjoin(b_parent, B.parent)
我獲得以下SQL:
SELECT ... FROM a JOIN b ON b.id = a.b_id LEFT OUTER JOIN b AS b_1 ON b_1.id = b.parent_id AND *b*.soft_deleted IS False
我期待下面的可是:
SELECT ... FROM a JOIN b ON b.id = a.b_id LEFT OUTER JOIN b AS b_1 ON b_1.id = b.parent_id AND *b_1*.soft_deleted IS False
當我寫有別名b查詢
def soft_delete_relationship(first, second, *args, **kwargs):
if isinstance(first, str) and isinstance(second, str):
other, other_column = first.split('.')
_this, this_column = second.split('.')
primaryjoin = ' & '.join(['({} == {})'.format(first, second), '{}.soft_deleted.is_(False)'.format(other)])
else:
other, other_column = args[0].split('.')
_this, this_column = args[1].split('.')
primaryjoin = lambda: (first == second) & getattr(second.table.c, 'soft_deleted').is_(False)
kwargs['primaryjoin'] = primaryjoin
return relationship(other, **kwargs)
出現的問題我明確寫下:
A.query.join(A.b).outerjoin(b_parent, (b_parent.id == B.parent_id) & b_parent.soft_deleted.is_(False))
我得到了正確的查詢。
如何在沒有顯式連接條件的情況下在查詢中獲得b_1的正確別名? 順便說一句,預計在SQLAlchemy 0.7.9中的SQL。