sqlalchemy查询一个与sqli的关系

Question

我已经编写了下面的示例代码来为我正在编写的ACL系统构建一个搜索索引。本例中的查询将重新返回分配了任何给定ACL的所有对象。但是我需要一个查询/过滤器来返回分配了所有acl的对象。在

感谢任何帮助。在

#!/usr/bin/env python
# -*- coding: utf-8 -*-
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship
from sqlalchemy.orm import sessionmaker
from sqlalchemy.orm import backref
from sqlalchemy import create_engine
from sqlalchemy import ForeignKey
from sqlalchemy import Integer
from sqlalchemy import String
from sqlalchemy import Column

_db_uri = "sqlite:////tmp/test.sql"
Base = declarative_base()
engine = create_engine(_db_uri, echo=False)
Session = sessionmaker(bind=engine)

class IndexObject(Base):
    """ Index object. """
    __tablename__ = 'objects'
    id = Column(Integer, primary_key=True)
    name = Column(String(128), unique=True, nullable=True)
    acls = relationship('IndexObjectACL',
                                cascade = "all,delete",
                                backref='objects',
                                lazy='dynamic')

    def __repr__(self):
        _repr_ =("<IndexObject (name='%s')>" % (self.name))
        return _repr_


class IndexObjectACL(Base):
    """ Index object ACL. """
    __tablename__ = 'acls'
    id = Column(Integer, primary_key=True)
    value = Column(String(128), nullable=False)
    oid = Column(Integer, ForeignKey('objects.id'))

    def __repr__(self):
        __repr__ = ("<IndexObjectACL (value='%s')>" % (self.value))
        return __repr__


object_list = [
        "object1",
        "object2",
        "object3",
    ]

acl_list = {
        "object1" : [
                    "view",
                    "edit",
                    "enable",
                    "delete",
                    ],
        "object2" : [
                    "view",
                    "edit",
                    ],
        "object3" : [
                    "enable",
                    "delete",
                    ],
    }


Base.metadata.create_all(engine)

session = Session()

for o in object_list:
    acls = []
    for acl in acl_list[o]:
        a = IndexObjectACL(value=acl)
        acls.append(a)

    index_object = IndexObject(name=o, acls=acls)
    session.add(index_object)

session.commit()


search_acls = [ "enable", "delete" ]
q = session.query(IndexObject)
q = q.join(IndexObject.acls).filter(IndexObjectACL.value.in_(search_acls))

print(q.all())

session.close()

Answer 1

queries = []
acl_q = q.join(IndexObject.acls)
for acl in search_acls:
    x = acl_q.filter(IndexObjectACL.value == acl)
    queries.append(x)
q = q.intersect(*queries)

我可以试着解释一下，但我对SQL炼金术和SQL还不太熟悉。所以我可能解释错了。。。join()根据IndexObject和IndexObjectACL表的关系将它们连接起来，这将导致一个新的查询。此查询用于为要使用filter()匹配的每个ACL创建一个新查询。最后，我们使用intersect()（SQL INTERSECT）来获取所有查询中出现的所有IndexObject。经过一些测试后，这似乎是一种快速搜索已分配了所有acl的对象的方法。它也很像Python。在

Answer 2

我认为这可能是一个使用division-的机会。IndexObjectACL除以searchACL应生成具有所有searchACL的IndexObject。换言之，对于不存在不在其IndexObjectACL中的searchACL的查询：

from sqlalchemy import union, select, literal

# Create an aliased UNION of all the search_acls to query against
search_acls_union = union(*(select([literal(acl).label('acl')])
                            for acl in search_acls)).alias()

# Query for those IndexObjects where...
# No SearchAcl exists where...
# No IndexObjectACL exists where value == acl AND oid == id
q = session.query(IndexObject).\
    filter(~session.query().select_from(search_acls_union).
           filter(~IndexObject.acls.any(value=search_acls_union.c.acl)).
           exists())

此查询的结果是

^{pr2}$

如果你加上

    "object4" : [
                "enable",
                ],
    "object5" : [
                "delete",
                ],

为了证明不返回部分匹配，它仍然只返回对象1和对象3。在

您最初的“have any”查询也可以重写为在SQL语言中使用semijoin，或{}：

q = session.query(IndexObject).\
    filter(IndexObject.acls.any(
        IndexObjectACL.value.in_(search_acls)))