如何按日期返回每个字段的最新行答案

【问题标题】：How to return latest rows by date for each field如何按日期返回每个字段的最新行
【发布时间】：2019-09-08 16:20:41
【问题描述】：

我在 postgres 9.4 上。我想查询我的adjustments 表以获取每个字段的最新调整。一些adjustments 可以调整多个字段。

示例 1

+---------------+-------------+----------------+---------------+
| available_qty | on_hand_qty |  backstock_qty |     date      |
+---------------+-------------+----------------+---------------+
| null          | 2           | null           | 2018-01-03    |
| 1             | null        | 10             | 2018-01-04    |
| null          | null        | 2              | 2018-01-01    |
| 3             | 3           | null           | 2018-01-02    |
+---------------+-------------+----------------+---------------+

...查询每个字段的最新调整将导致：

+---------------+-------------+----------------+---------------+
| available_qty | on_hand_qty |  backstock_qty |     date      |
+---------------+-------------+----------------+---------------+
| null          | 2           | null           | 2018-01-03    |
| 1             | null        | 10             | 2018-01-04    |
+---------------+-------------+----------------+---------------+

示例 2

+---------------+-------------+----------------+---------------+
| available_qty | on_hand_qty |  backstock_qty |     date      |
+---------------+-------------+----------------+---------------+
| null          | 2           | null           | 2018-01-03    |
| 1             | null        | null           | 2018-01-04    |
| null          | null        | 2              | 2018-01-01    |
| null          | null        | 3              | 2018-01-02    |
| 3             | null        | null           | 2018-01-02    |
| null          | 4           | null           | 2018-01-01    |
+---------------+-------------+----------------+---------------+

...查询每个字段的最新调整将导致：

+---------------+-------------+----------------+---------------+
| available_qty | on_hand_qty |  backstock_qty |     date      |
+---------------+-------------+----------------+---------------+
| null          | 2           | null           | 2018-01-03    |
| 1             | null        | null           | 2018-01-04    |
| null          | null        | 3              | 2018-01-02    |
+---------------+-------------+----------------+---------------+

我可以通过大量UNION 查询来实现这一点：

(SELECT * FROM adjustments WHERE available_qty IS NOT NULL ORDER BY date DESC LIMIT 1)
UNION
(SELECT * FROM adjustments WHERE on_hand_qty IS NOT NULL ORDER BY date DESC LIMIT 1)
UNION
(SELECT * FROM adjustments WHERE backstock_qty IS NOT NULL ORDER BY date DESC LIMIT 1)

...但是这可以通过一个查询来实现吗？实际上，这里可以有很多领域。提前致谢！

【问题讨论】：

试试这个：stackoverflow.com/a/20748872/11610186，在你的情况下：sqlfiddle.com/#!17/f1a44/5
@GrzegorzSkibinski 谢谢，快到了！有没有办法同时返回每个对应字段的日期值？
理想情况下，我希望返回实际的 adjustments 行
我把整个事情放在下面的答案中。

标签： sql postgresql postgresql-9.4

【解决方案1】：

好的，这就是您要查找的内容：（顺序是 - 字段，然后是最近调整的日期）

select 
   first_value(available_qty) over (order by available_qty is null, date desc), 
   first_value(date) over (order by available_qty is null, date desc), 
   first_value(on_hand_qty) over (order by on_hand_qty is null, date desc),
   first_value(date) over (order by on_hand_qty is null, date desc),
   first_value(backstock_qty) over (order by backstock_qty is null, date desc),
   first_value(date) over (order by backstock_qty is null, date desc)
from
   adjustment
limit 1

并更新SQL fiddle

【讨论】：