SQL

Question

我的 table 大得多，但剪短一点会像：

---------+---+----------+--------+------------+---
|distance|qtt|deliver_by| store  |deliver_time| ...
+--------+---+----------+--------+------------|---
|   11   |  1|  pa      | store_a|  1111      |
|   123  |  2|  pa      | store_a|  1112      |
|   33   |  3|  pb      | store_a|  1113      |
|   33   |  2|  pa      | store_b|  2221      |
|   44   |  2|  pb      | store_b|  2222      |
|   5    |  2|  pc      | store_b|  2223      |
|   5    |  2|  pc      | store_b|  2224      |
|   6    |  5|  pb      | store_c|  3331      |
|   7    |  5|  pb      | store_c|  3332      |
----------------------------------------------....

有多家商店，但只有 3 家可能送货（deliver_by：pa、pb 和 pc) 在特定时间交付产品。考虑 deliver_time 一个时间戳。

我想 select 整个 table 和 添加 6 个新列, min 和 max 次在商店中每 deliver_by 次。商店可以由 3 种交付方式（pa、pb、pc）中的任何一种提供服务，但不是必需的。

我几乎可以完成正确的结果，使用下面的查询，问题是在案例 deliver_by pX 不存在，我没有得到 null 而是 min/max到店送货。

我真的很想使用分区依据，所以我写了这个来添加新的 min/max 列：

select
  min(deliver_time) over (partition by store, deliver_by='pa') as as min_time_sd_pa
, max(deliver_time) over (partition by store, deliver_by='pa') as as min_time_sd_pa

, min(deliver_time) over (partition by store, deliver_by='pb') as as min_time_sd_pb
, max(deliver_time) over (partition by store, deliver_by='pb') as as min_time_sd_pb

, min(deliver_time) over (partition by store, deliver_by='pc') as as min_time_sd_pc
, max(deliver_time) over (partition by store, deliver_by='pc') as as min_time_sd_pc

, distance, qtt, ....
from mytable

正确的输出应该是:

min_time_sd_pa|max_time_sd_pa|min_time_sd_pb|max_time_sd_pb|min_time_sd_pc|max_time_sd_pc|distance|qtt|deliver_by| store  |deliver_time
--------------+--------------+--------------+--------------+--------------+--------------+--------+---+----------+--------+------------
    1111      |  1112        |  1113        |   1113       |   null       |   null       |   11   |  1| pa       | store_a| 1111
    1111      |  1112        |  1113        |   1113       |   null       |   null       |   123  |  2| pa       | store_a| 1112
    1111      |  1112        |  1113        |   1113       |   null       |   null       |   33   |  3| pb       | store_a| 1113
    2221      |  2221        |  2222        |   2222       |   2223       |   2224       |   33   |  2| pa       | store_b| 2221
    2221      |  2221        |  2222        |   2222       |   2223       |   2224       |   44   |  2| pb       | store_b| 2222
    2221      |  2221        |  2222        |   2222       |   2223       |   2224       |   5    |  2| pc       | store_b| 2223
    2221      |  2221        |  2222        |   2222       |   2223       |   2224       |   5    |  2| pc       | store_b| 2224
    null      |  null        |  null        |   null       |   3331       |   3332       |   6    |  5| pb       | store_c| 3331
    null      |  null        |  null        |   null       |   3331       |   3332       |   7    |  5| pb       | store_c| 3332
---------------------------------------------------------------------------------------------------------------------------------------

我的 select min(..) over.. 语句中缺少什么，或者我怎样才能以最简单的方式实现这个结果？我正在使用 Hive QL，但我想这在大多数 SQL DBMS.

中都是通用的

谢谢

Answer 1

您可以在 min 和 max 中使用 case 表达式。

select
 min(case when deliver_by='pa' then deliver_time end) over (partition by store) as min_time_sd_pa
,max(case when deliver_by='pa' then deliver_time end) over (partition by store) as max_time_sd_pa
,min(case when deliver_by='pb' then deliver_time end) over (partition by store) as min_time_sd_pb
,max(case when deliver_by='pb' then deliver_time end) over (partition by store) as max_time_sd_pb
,min(case when deliver_by='pc' then deliver_time end) over (partition by store) as min_time_sd_pc
,max(case when deliver_by='pc' then deliver_time end) over (partition by store) as max_time_sd_pc
,m.*
from mytable m

SQL - 按一列和一些字段类型分区

SQL - partition by one column and some field types

hive

window

partition