Question

我有以下SQLAlchemy映射类：

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user.email = "user@email.com"获取这样的表：

email | name | document_name | document_readAllowed | document_writeAllowed

如何使用SQLAlchemy的一个查询请求？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，

Answer 1

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())

Answer 2

一个好的风格是设置一些关系和一个主键用于权限（实际上，通常情况下，为所有内容设置整数主键是好的方式，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')

然后用连接做一个简单的查询：

query = session.query(User, Document, DocumentsPermissions).join(Document).join(DocumentsPermissions)

Answer 3

正如@letitbee所说，最佳做法是将主键分配给表并正确定义关系以允许正确的ORM查询。那就是说......

如果您有兴趣按以下方式编写查询：

SELECT
    user.email,
    user.name,
    document.name,
    documents_permissions.readAllowed,
    documents_permissions.writeAllowed
FROM
    user, document, documents_permissions
WHERE
    user.email = "user@email.com";

然后你应该去寻找类似的东西：

session.query(
    User, 
    Document, 
    DocumentsPermissions
).filter(
    User.email == Document.author
).filter(
    Document.name == DocumentsPermissions.document
).filter(
    User.email == "user@email.com"
).all()

相反，如果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

然后你应该按照以下方式做点什么：

session.query(
    User
).join(
    Document
).join(
    DocumentsPermissions
).filter(
    User.email == "user@email.com"
).all()

关于这一点......

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

有关详细信息，请访问docs。

Answer 4

扩展Abdul的回答，您可以通过加入列来获得KeyedTuple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()

Answer 5

此函数将生成必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)

sqlalchemy：如何通过一个查询连接多个表？

5 个答案: