Python 在SQLAlchemy关系中放置子类实例
TLDR;问题在于继承构造,我不知道在没有声明性API的情况下如何构造继承 我创建了通用模型Python 在SQLAlchemy关系中放置子类实例,python,sqlalchemy,Python,Sqlalchemy,TLDR;问题在于继承构造,我不知道在没有声明性API的情况下如何构造继承 我创建了通用模型Job,它将通过DeploymentJob这样的子类进一步缩小范围。每个作业由几个操作组成。如果我定义了这个JobAction关系,我就不能在Job子类实例中使用它 from sqlalchemy import (Table, Binary, Column as C, String, Integer, ForeignKey, create_engine, M
Job
,它将通过DeploymentJob
这样的子类进一步缩小范围。每个作业
由几个操作组成
。如果我定义了这个JobAction
关系,我就不能在Job
子类实例中使用它
from sqlalchemy import (Table, Binary, Column as C, String, Integer,
ForeignKey, create_engine, MetaData)
from sqlalchemy.orm import (mapper, relationship, backref, scoped_session,
sessionmaker)
metadata = MetaData()
db_engine = create_engine('sqlite:////tmp/test.db', convert_unicode=True)
db_session = scoped_session(sessionmaker(bind=db_engine))
class Job(object):
pass
class DeploymentJob(Job):
def __init__(self, *args, **kwargs):
super(DeploymentJob, self).__init__(*args, **kwargs)
class Action(object):
def __init__(self, unit, job):
self.unit = unit
self.job = job
jobs = Table('jobs', metadata,
C('id', Integer, primary_key=True),
C('type', String, nullable=False)
)
deployment_jobs = Table('deployment_jobs', metadata,
C('id', ForeignKey('jobs.id'), primary_key=True)
)
actions = Table('actions', metadata,
C('job_id', ForeignKey('jobs.id'), primary_key=True),
C('unit', String, primary_key=True)
)
mapper(Job,
jobs,
polymorphic_on=jobs.c.type,
properties = {
'actions': relationship(Action, lazy='dynamic', uselist=True,
backref=backref('job', uselist=False)),
}
)
mapper(DeploymentJob, deployment_jobs, polymorphic_identity='deployment')
mapper(Action, actions)
metadata.create_all(bind=db_engine)
unit = 'second-machine'
job = DeploymentJob()
action = Action(unit, job)
print "action.job -> %s is job: %s" % (action.job, isinstance(action.job, Job))
# >> action.job -> <__main__.DeploymentJob object at 0x7fe> is job: True
db_session.add(action)
db_session.add(job)
db_session.commit()
问题是
Mapper
不会将DeploymentJob
作为Job
的子对象计算,直到Job
的Mapper对象作为inherit
参数提供给DeploymentJob
的Mapper对象。所以这是可行的:
JobsMapping = mapper(Job,
jobs,
polymorphic_on=jobs.c.type,
properties = {
'actions': relationship(Action, lazy='dynamic', uselist=True,
backref=backref('job', uselist=False)),
}
)
mapper(DeploymentJob, deployment_jobs, inherits=JobsMapping, polymorphic_identity='deployment')
JobsMapping = mapper(Job,
jobs,
polymorphic_on=jobs.c.type,
properties = {
'actions': relationship(Action, lazy='dynamic', uselist=True,
backref=backref('job', uselist=False)),
}
)
mapper(DeploymentJob, deployment_jobs, inherits=JobsMapping, polymorphic_identity='deployment')