Question

我在SQLAlchemy中有一个类，它与同一个辅助表有多个关系。看起来有点像这样：

class Job(Base):
    __tablename__ = 'jobs'
    id = Column(Integer, primary_key=True)
    tasks_queued = relationship("Task", lazy="dynamic",
        primaryjoin="(Task.state == 'queued') & (Task.job_id == Job.id)")
    tasks_running = relationship("Task", lazy="dynamic",
        primaryjoin="(Task.state == 'running') & (Task.job_id == Job.id)")
    tasks_done = relationship("Task", lazy="dynamic",
        primaryjoin="(Task.state == 'done') & (Task.job_id == Job.id)")
    tasks_failed = relationship("Task", lazy="dynamic",
        primaryjoin="(Task.state == 'failed') & (Task.job_id == Job.id)")

class Task(Base):
    __tablename__ = 'tasks'
    id = Column(Integer, primary_key=True)
    job_id = Column(Integer, ForeignKey("jobs.id"))
    state = Column(String(8), nullable=False, default='queued')
    job = relationship("Job")

作业有零个或多个任务。任务可以具有四种状态之一：＆＃34;排队＆＃34;，＆＃34;运行＆＃34;，＆＃34;完成＆＃34;或＆＃34;失败＆＃34;。在查询作业时，我希望看到按状态分割的那些任务的计数，即每个作业分别有多少个排队，运行，完成和失败的任务。我还希望能够通过任何这些计数对输出进行排序。

经过一段谷歌搜索后，我发现了如何为一种关系做到这一点：

session.query(Job, func.count(Job.tasks_queued).label("t_queued")).\
outerjoin(Job.tasks_queued).group_by(Job).order_by("t_queued ASC").all()

但是，一旦我尝试将其扩展到多个关系，事情就会变得模糊：

session.query(Job, func.count(Job.tasks_queued).label("t_queued"), 
    func.count(Job.tasks_running).label("t_running")).\
outerjoin(Job.tasks_queued).\
outerjoin(Job.tasks_running).group_by(Job).order_by("t_queued ASC").all()

产生此错误：

sqlalchemy.exc.OperationalError: (OperationalError) ambiguous column name: tasks.state 'SELECT jobs.id AS jobs_id, count(tasks.state = ? AND tasks.job_id = jobs.id) AS t_queued, count(tasks.state = ? AND tasks.job_id = jobs.id) AS t_running \nFROM jobs LEFT OUTER JOIN tasks ON tasks.state = ? AND tasks.job_id = jobs.id LEFT OUTER JOIN tasks ON tasks.state = ? AND tasks.job_id = jobs.id GROUP BY jobs.id ORDER BY t_queued ASC' ('queued', 'running', 'queued', 'running')

所以我需要告诉sqlalchemy第一个计数是指第一个连接，第二个计数是指第二个连接。在纯SQL中，我只是给连接表ad-hoc别名，然后引用那些别名而不是count（）函数中的表名。我如何在SQLAlchemy中执行此操作？

Answer 1

您可以将aliases与sqlalchemy：

一起使用

a_q = aliased(Task)
a_r = aliased(Task)
a_d = aliased(Task)
a_f = aliased(Task)
qry2 = (session.query(Job,
                      func.count(a_q.id.distinct()).label("t_queued"),
                      func.count(a_r.id.distinct()).label("t_running"),
                      func.count(a_d.id.distinct()).label("t_done"),
                      func.count(a_f.id.distinct()).label("t_failed"),
                      )
        .outerjoin(a_q, Job.tasks_queued)
        .outerjoin(a_r, Job.tasks_running)
        .outerjoin(a_d, Job.tasks_done)
        .outerjoin(a_f, Job.tasks_failed)
        .group_by(Job)
        .order_by("t_queued ASC")

我认为您需要向distinct添加count。

SQLAlchemy：通过多个关系对同一个表进行计数和排序

1 个答案: