ORDER BY取决于两个日期列

Question

我正在使用Postgresql 10，并尝试对我的选择查询进行排序，以使我拥有的两个日期列之间的较小日期排在最前面。

这里是sqlfiddle

排序结果将如下所示：

+----+--------------+--------------+
| id | column1_date | column2_date |
+----+--------------+--------------+
| 1  | 2018-05-07   | null         |
+----+--------------+--------------+
| 3  | null         | 2018-05-28   |
+----+--------------+--------------+
| 2  | 2018-06-28   | 2018-06-30   |
+----+--------------+--------------+
| 4  | null         | 2018-07-12   |
+----+--------------+--------------+

我试图这样做：

SELECT * FROM Test
    ORDER BY 
      CASE WHEN column1_date is null AND column2_date is not null
          THEN column2_date END ASC, column1_date ASC,

      CASE WHEN column1_date is not null AND column2_date is null
          THEN column1_date END ASC, column2_date ASC,

      CASE WHEN column1_date is not null AND column2_date is not null AND column1_date < column2_date
          THEN column1_date END ASC, column2_date ASC,

      CASE WHEN column1_date is not null AND column2_date is not null AND column1_date > column2_date
          THEN column2_date END ASC, column1_date ASC

但这并不成功。 我想念什么？

谢谢

Answer 1

您可以使用

ORDER BY least(coalesce(column1_date, DATE 'infinity'),
               coalesce(column2_date, DATE 'infinity'))

这将始终使用两者中较小的一个。

如果两列均为NULL，则该行将排在最后。

Answer 2

您可以使用内部子查询：

select 
  t.id, t.column1_date, t.column2_date
from (
  select 
    id, 
    column1_date, 
    column2_date,
    CASE 
      WHEN column1_date is null then column2_date
      WHEN column2_date is null then column1_date
      WHEN column1_date < column2_date THEN column1_date
      ELSE column2_date
    END AS min_date
  from Test) as t
order by t.min_date;