我有以下SQLAlchemy映射的类:
class User(Base):
__tablename__ = 'users'
email = Column(String, primary_key=True)
name = Column(String)
class Document(Base):
__tablename__ = "documents"
name = Column(String, primary_key=True)
author = Column(String, ForeignKey("users.email"))
class DocumentsPermissions(Base):
__tablename__ = "documents_permissions"
readAllowed = Column(Boolean)
writeAllowed = Column(Boolean)
document = Column(String, ForeignKey("documents.name"))
我需要为这样的一张桌子user.email = "user@email.com"
:
email | name | document_name | document_readAllowed | document_writeAllowed
如何使用一个对SQLAlchemy的查询请求来完成?以下代码对我不起作用:
result = session.query(User, Document, DocumentPermission).filter_by(email = "user@email.com").all()
谢谢,
试试这个
q = Session.query(
User, Document, DocumentPermissions,
).filter(
User.email == Document.author,
).filter(
Document.name == DocumentPermissions.document,
).filter(
User.email == 'someemail',
).all()
一种好的样式是为权限设置一些关系和一个主键(实际上,通常为所有的东西设置整数主键是一种好样式,但无论如何):
class User(Base):
__tablename__ = 'users'
email = Column(String, primary_key=True)
name = Column(String)
class Document(Base):
__tablename__ = "documents"
name = Column(String, primary_key=True)
author_email = Column(String, ForeignKey("users.email"))
author = relation(User, backref='documents')
class DocumentsPermissions(Base):
__tablename__ = "documents_permissions"
id = Column(Integer, primary_key=True)
readAllowed = Column(Boolean)
writeAllowed = Column(Boolean)
document_name = Column(String, ForeignKey("documents.name"))
document = relation(Document, backref = 'permissions')
然后使用联接进行简单查询:
query = session.query(User, Document, DocumentsPermissions).join(Document).join(DocumentsPermissions)
正如@letitbee所说,最佳做法是为表分配主键并正确定义关系,以进行正确的ORM查询。话虽如此...
如果您有兴趣通过以下方式编写查询:
SELECT
user.email,
user.name,
document.name,
documents_permissions.readAllowed,
documents_permissions.writeAllowed
FROM
user, document, documents_permissions
WHERE
user.email = "user@email.com";
然后,您应该尝试类似:
session.query(
User,
Document,
DocumentsPermissions
).filter(
User.email == Document.author
).filter(
Document.name == DocumentsPermissions.document
).filter(
User.email == "user@email.com"
).all()
如果相反,您想要执行以下操作:
SELECT 'all the columns'
FROM user
JOIN document ON document.author_id = user.id AND document.author == User.email
JOIN document_permissions ON document_permissions.document_id = document.id AND document_permissions.document = document.name
然后,您应该按照以下方式进行操作:
session.query(
User
).join(
Document
).join(
DocumentsPermissions
).filter(
User.email == "user@email.com"
).all()
关于这一点的说明...
query.join(Address, User.id==Address.user_id) # explicit condition
query.join(User.addresses) # specify relationship from left to right
query.join(Address, User.addresses) # same, with explicit target
query.join('addresses') # same, using a string
有关更多信息,请访问docs。
扩展Abdul的答案,您可以KeyedTuple
通过连接列来获得行而不是离散的行集合:
q = Session.query(*User.__table__.columns + Document.__table__.columns).\
select_from(User).\
join(Document, User.email == Document.author).\
filter(User.email == 'someemail').all()
此函数将生成所需的表作为元组列表。
def get_documents_by_user_email(email):
query = session.query(
User.email,
User.name,
Document.name,
DocumentsPermissions.readAllowed,
DocumentsPermissions.writeAllowed,
)
join_query = query.join(Document).join(DocumentsPermissions)
return join_query.filter(User.email == email).all()
user_docs = get_documents_by_user_email(email)
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文章标签:join , python , sql , sqlalchemy
版权声明:本文为原创文章,版权归 admin 所有,欢迎分享本文,转载请保留出处!
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