如何在postgresql中实现与same table的匹配？

Question

我想根据以下标准计算按国家划分的网站数量：-

对于 Trial_Start_Date = X 和 Site_Activated = Y 的每个 TrialSite，您应该计算满足这些条件的所有行：

Trial_Start_Date <= Y, AND
TrialSite_Activation_Date >= X 即所有与该行的试用开始日期到 TrialSite 激活日期有重叠时间段的行。

示例数据示例：

Trial id  start_date  country  site_id trialSite_activation_date

Trial A    01-01-2017  India  site1 01-02-2017 ----> 2 (only overlaps with itself, and with 2nd row)

Trial A  01-01-2017  India  site 2 01-04-2017 ----> 4 (overlaps with all rows, including itself)

Trial B    02-03-2017 India  site3 01-04-2017 ----> 3 (does not overlap with first row, since Trial_Start_Date > 01-02-2017)

Trial B   02-03-2017 India  site4 01-04-2017 ----> 3

此数据可以包含多个国家，此逻辑需要应用于相同的国家记录。

Answer 1

您可以在 daterange 上使用“重叠”运算符 &&：

SELECT t1, count(*)
FROM trial t1
   JOIN trial t2
      ON daterange(t1.trial_start_date, t1.trialsite_activation_date, '[]')
         && daterange(t2.trial_start_date, t2.trialsite_activation_date, '[]')
GROUP BY t1;

                      t1                       | count
-----------------------------------------------+-------
 ("Trial A",2017-01-01,India,site1,2017-02-01) |     2
 ("Trial A",2017-01-01,India,site2,2017-04-01) |     4
 ("Trial B",2017-03-02,India,site3,2017-04-01) |     3
 ("Trial B",2017-03-02,India,site4,2017-04-01) |     3
(4 rows)

而不是在 SELECT 列表中使用整行引用 t1，您可以在那里指定单独的列，但随后您将它们列在 GROUP BY 子句中作为嗯。

如何在postgresql中实现与same table的匹配？

How to achieve matching with same table in postgresql?

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