我正在尝试使用OUTER JOIN执行relationship()
,以便在加入第二个表时加入第二个表。我目前仍然坚持如何做到这一点,我似乎无法找出options()
,relationship()
和outerjoin()
的正确组合。
我有以下表格,如果存在具有应用程序ID和artistID(由函数提供)的行,我正在尝试将AppLike加入到应用程序中
很高兴提供任何其他信息,我已经有一个我的联接工作,如下所示,但总会有一行匹配。
from sqlalchemy import Column
from . import Base
from . import DBSession
from sqlalchemy.dialects.mysql import (
INTEGER,
VARCHAR,
TEXT,
TINYINT,
)
from sqlalchemy.sql import and_
from sqlalchemy import ForeignKey
from sqlalchemy.orm import relationship, joinedload
import time
# 0 = new
# 1 = Denied
# 2 = Accepted
def getNewApplications(artistID):
query = DBSession.query(Application).\
options(joinedload('pieces')).\
options(joinedload('vote')).\
filter(AppLike.artist_id==artistID).\
filter(Application.approved==0)
#join(AppPiece, Application.app_id==AppPiece.app_id).\
#outerjoin(AppLike, and_(Application.app_id==AppLike.app_id,
# AppLike.artist_id==artistID)).\
import pdb; pdb.set_trace()
return query.all()
class Application(Base):
""" The SQLAlchemy declarative model class for a FileFavorite object. """
__tablename__ = 'applications'
__table_args__ = {
'mysql_engine': 'InnoDB',
'mysql_charset': 'utf8'
}
app_id = Column(INTEGER(11), autoincrement=True, primary_key=True, nullable=False)
name = Column(VARCHAR(64), nullable=False)
nickname = Column(VARCHAR(64), nullable=False)
email = Column(VARCHAR(255), nullable=False)
description = Column(TEXT(), nullable=False)
profile_link = Column(VARCHAR(128), nullable=False)
location = Column(VARCHAR(64), nullable=False)
approved = Column(TINYINT(4), nullable=False)
pieces = relationship("AppPiece", lazy='joined')
vote = relationship("AppLike", lazy='joined')
def __init__(self, name, nickname, email, desc, profileLink,
location, approved):
self.name = name
self.nickname = nickname
self.email = email
self.description = desc
self.profile_link = profileLink
self.location = location
self.approved = approved
class AppPiece(Base):
""" The SQLAlchemy declarative model class for a FileFavorite object. """
__tablename__ = 'app_pieces'
__table_args__ = {
'mysql_engine': 'InnoDB',
'mysql_charset': 'utf8'
}
app_piece_id = Column(INTEGER(11), autoincrement=True, primary_key=True, nullable=False)
app_id = Column(INTEGER(11), ForeignKey('applications.app_id'))
link = Column(VARCHAR(128), nullable=False)
def __init__(self, appID, link):
self.app_id = appID
self.link = link
class AppLike(Base):
""" The SQLAlchemy declarative model class for a FileFavorite object. """
__tablename__ = 'app_likes'
__table_args__ = {
'mysql_engine': 'InnoDB',
'mysql_charset': 'utf8'
}
app_id = Column(INTEGER(11), ForeignKey('applications.app_id'))
artist_id = Column(INTEGER(11), primary_key=True, nullable=False)
vote = Column(TINYINT(4), nullable=False)
def __init__(self, appID, artistID, vote):
self.app_id = appID
self.artist_id = artistID
self.vote = vote
答案 0 :(得分:2)
您绝对不需要options(joinedload('pieces'))
,它已在您的模型中定义(lazy='joined'
)。连接条件在这里是棘手的部分,需要使用子查询来完成,因为我们也希望在那里进行过滤。因此,最终查询应如下所示:
# We do the filtering on AppLike in the subquery and later join
# Application to it.
applike_subq = DBSession.query(AppLike).\
filter(AppLike.artist_id == artistID).subquery()
query = DBSession.query(Application).\
outerjoin(applike_subq, Application.vote).\
filter(Application.approved == 0).all()