[英]Many to many query in sqlalchemy
我的问题有表格。
class TemplateExtra(ExtraBase, InsertMixin, TimestampMixin):
__tablename__ = 'template_extra'
id = Column(Integer, primary_key=True, autoincrement=False)
name = Column(Text, nullable=False)
roles = relationship(
'RecipientRoleExtra',
secondary='template_to_role',
)
class RecipientRoleExtra(
ExtraBase, InsertMixin, TimestampMixin,
SelectMixin, UpdateMixin,
):
__tablename__ = 'recipient_role'
id = Column(Integer, primary_key=True, autoincrement=True)
name = Column(Text, nullable=False)
description = Column(Text, nullable=False)
class TemplateToRecipientRoleExtra(ExtraBase, InsertMixin, TimestampMixin):
__tablename__ = 'template_to_role'
id = Column(Integer, primary_key=True, autoincrement=True)
template_id = Column(Integer, ForeignKey('template_extra.id'))
role_id = Column(Integer, ForeignKey('recipient_role.id'))
我想在两个 sql 查询中选择所有具有预取角色的模板,就像 Django ORM 对 prefetch_related 所做的那样。 我可以做吗? 这是我目前的尝试。
def test_custom():
# creating engine with echo=True
s = DBSession()
for t in s.query(TemplateExtra).join(RecipientRoleExtra, TemplateExtra.roles).all():
print(f'id = {t.id}')
for r in t.roles:
print(f'-- {r.name}')
但..
FROM recipient_role, template_to_role
WHERE %(param_1)s = template_to_role.template_id AND recipient_role.id = template_to_role.role_id
FROM recipient_role, template_to_role
with WHERE %(param_1)s = template_to_role.template_id AND recipient_role.id = template_to_role.role_id
。 这是正确的吗?你能帮我吗?
基于这个答案: flask many to many join as done by prefetch_related from django
也许是这样的:
roles = TemplateExtra.query.options(db.joinedload(TemplateExtra.roles)).all
让我知道它是否有效。
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