如何为 SQL 中的每个特定列应用窗口？ [复制]

Posted 2023-02-16

【中文标题】如何为 SQL 中的每个特定列应用窗口？ [复制]

【英文标题】：How to apply windows for each specific column in SQL? [duplicate]

【发布时间】：2021-06-13 17:25:42

【问题描述】：

我想为每个特定的 unit_of_measure 每个用户完成最后一个事件：

我有这张桌子：

person_id   event_time       event_derscription   unit_of_measure 
-----------------------------------------------------------------
1           20200801120101  "some description"     "unit1"
1           20200801120501  "some description 2"   "unit1"
1           20200801120501  "some description 2"   "unit9"
2           20200801120301  "some description 3"   "unit1"
2           20200801120501  "some description 4"   "unit1"

预期输出是：

person_id   event_time       event_derscription   unit_of_measure 
-----------------------------------------------------------------
1           20200801120101  "some description"     "unit1"
2           20200801120301  "some description 2"   "unit1"
1           20200801120501  "some description 2"   "unit9"

我尝试了什么：

select * 
from 
    (select 
         person_id, event_time, event_derscription, unit_of_measure, 
         rank() over (partition by unit_of_measure order by event_time desc) as RN 
     from 
         test.person_events 
     where 
         partition_name = 20200801 
     group by 
         person_id, event_time, event_description, unit_of_measure) 
where 
    RN = 1;  // I try to use group by person_id to get the result for each person_id but it did not work 

我上面代码的输出是：

person_id   event_time       event_derscription   unit_of_measure 
-----------------------------------------------------------------
2           20200801120301  "some description 2"   "unit1"
1           20200801120501  "some description 2"   "unit9"

我做错了什么吗？

【参考方案1】：

我认为你想要的查询是：

select person_id, event_time, event_derscription, unit_of_measure
from (select pe,
             row_number() over (partition BY unit_of_measure, person_id order by event_time desc) as seqnum
      from test.person_events pe
      where partition_name = 20200801 
     ) pe
where seqnum = 1; 

注意事项：

解决问题的主要方法是将person_id 包含在partition by 中。 我认为不需要group by。您的问题中没有任何内容提到它为什么是可取的。 要获取一行，请使用row_number() 而不是rank()。即使你没有重复，它也传达了你想要一排的意图。

【讨论】：

@code 。 . .我现在意识到您故意使用rank() 来获取不同的person_ids，然后使用group by 将其删除。巧妙地尝试将各个部分组合在一起——但这是要走的路。