如何在Django ORM中表示复杂的子查询和计算表？

Question

我有以下基本可行的SQL。我将如何在Django ORM中表示这一点？我想避免运行完整的原始查询

我不知道如何在Django ORM中进行子查询以及如何正确执行笛卡尔积（由CROSS JOIN实现）

SELECT datum,
       alldata.worker_id,
       reporting_plan.project_id,
       SUM(effort::float)/60/60
FROM
  (SELECT DISTINCT datum,
                   reporting_plan.worker_id AS worker_id
   FROM
     (SELECT datum::date
      FROM generate_series('2019-05-01', '2019-12-31', '1 day'::interval) datum) AS dates
   CROSS JOIN reporting_plan
   ORDER BY datum,
            worker_id) AS alldata
LEFT OUTER JOIN reporting_plan ON alldata.worker_id = reporting_plan.worker_id
AND datum <= reporting_plan.end
AND datum >= reporting_plan.start
GROUP BY datum,
         alldata.worker_id,
         reporting_plan.worker_id,
         reporting_plan.project_id
ORDER BY datum,
         alldata.worker_id,
         reporting_plan.worker_id,
         reporting_plan.project_id

预期结果是一个列表，其中包含时间范围内的所有日期以及所有工人和匹配的计划信息（项目和工作量）。

谢谢！

编辑：

基于@jimjimjim的反馈，我设法删除了CROSS JOIN，同时获得了相同的结果：

SELECT datum::date,
       alldata.worker_id,
       reporting_plan.project_id,
       SUM(effort::float)/60/60
FROM
  (SELECT DISTINCT generate_series ('2019-05-01', '2019-12-31', '1 day'::interval) AS datum,
                   worker_id
   FROM reporting_plan
   ORDER BY datum,
            worker_id) AS alldata
LEFT OUTER JOIN reporting_plan ON alldata.worker_id = reporting_plan.worker_id
AND datum <= reporting_plan.end
AND datum >= reporting_plan.start
GROUP BY datum,
         alldata.worker_id,
         reporting_plan.worker_id,
         reporting_plan.project_id
ORDER BY datum,
         alldata.worker_id,
         reporting_plan.project_id

Answer 1

这里你的问题与Reddit有点不同，用这种方法你将进行原始查询但返回ORM对象。我会做以下事情：

1. 为查询创建模型：

    class WorkerEffort(models.Model):
        date = models.DateField(...)
        worker = models.ForeignKey(Worker, db_column="worker_id")
        project = models.ForeignKey(Project, db_column="project_id")
        effort = models.DecimalField()

        class Meta:
            managed = False

管理信息：https://docs.djangoproject.com/en/2.2/ref/models/options/#managed

2. 对该模型执行raw查询：

WorkerEffort.objects.raw('''
SELECT datum,
       alldata.worker_id,
       reporting_plan.project_id,
       SUM(effort::float)/60/60
FROM
  (SELECT DISTINCT datum,
                   worker_id
   FROM
     (select *, generate_series(start, end, '1 day'::interval) as datum from reporting_plan) 
     ) AS alldata
LEFT OUTER JOIN reporting_plan ON alldata.worker_id = reporting_plan.worker_id
GROUP BY datum,
         alldata.worker_id,
         reporting_plan.worker_id,
         reporting_plan.project_id
ORDER BY datum,
         alldata.worker_id,
         reporting_plan.worker_id,
         reporting_plan.project_id
''')

使用原始查询，您将能够通过WorkerEffort（查询模型）将ID与ORM对象进行匹配。

Answer 2

这会给你相同的结果吗？它消除了交叉连接，这使得使用ORM更容易重写，并且由于系列是使用reporting_plan.start和end生成的，因此您不需要在连接条件中包含它们。如果查询中给出的日期需要参数化，则可以将它们作为参数包含在ORM命令中，而不是查询本身。

SELECT datum,
       alldata.worker_id,
       reporting_plan.project_id,
       SUM(effort::float)/60/60
FROM
  (SELECT DISTINCT datum,
                   worker_id
   FROM
     (select *, generate_series(start, end, '1 day'::interval) as datum from reporting_plan) 
     ) AS alldata
LEFT OUTER JOIN reporting_plan ON alldata.worker_id = reporting_plan.worker_id
GROUP BY datum,
         alldata.worker_id,
         reporting_plan.worker_id,
         reporting_plan.project_id
ORDER BY datum,
         alldata.worker_id,
         reporting_plan.worker_id,
         reporting_plan.project_id