我如何在 Django ORM 中表示复杂的子查询和计算表？

Question

我有以下基本有效的 SQL。我将如何在 Django ORM 中表示它？我想避免运行完整的原始查询

我不确定如何处理 Django ORM 中的子查询以及如何正确执行笛卡尔积（通过 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和所有工作人员以及匹配的计划信息（项目和工作量）。

谢谢！

编辑:

根据@jimjimjim 的反馈，我设法删除了 CROSS JOIN，但得到了相同的结果：

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

这给你同样的结果吗？它消除了交叉连接，这将使使用 ORM 重写变得更容易，并且由于该系列是使用 reporting_plan.start 和 end 生成的，因此您不需要将它们包含在连接条件中.如果查询中给出的日期需要参数化，您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

Answer 2

你在这里的问题与 Reddit 上的问题有点不同，使用这种方法你将进行原始查询但返回 ORM 对象。我将执行以下操作：

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

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

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 对象匹配。

我如何在 Django ORM 中表示复杂的子查询和计算表？

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django

django-orm