Question

我有这样的情况

~id ~from ~to ~label ~weight
100  A     B   knows    2
100  A     B   knows    3
100  A     B   knows    4

但是我只想要最大日期的权重。如何修改下面的CASE语句，以使ID仅存在1个条目。

查询：

(
select distinct 
CASE WHEN *some-condition* as "~id" 
,CASE *some-condition* as "~from" 
,CASE *some-condition* as "~to"
,CASE *some-condition* as "~label"
,CASE ??? as "weight"
from 
(select 
dense_rank() over(partition by t.job_id order by start_time desc) rnk,                                 
t.Date,
t.job_id,
t.start_time,
t.end_time,
t.dep_id,
t.table_name
.....
t.region_id,
from Table1 t
    ,Tabel2 J  
where t.JOB_ID=J.JOB_ID
)
where rnk=1
order by JOB_ID,table_name
) 
where "~id" is NOT NULL and "~label" is NOT NULL and "~from" is NOT NULL and "~to" is NOT NULL;
;

表t

job_id    Date      table_name .......   dep_id   weight
100     2020-10-20     abc                  1        2
100     2020-10-20     abc                  2        3
100     2020-10-20     abc                  3        4
100     2020-10-20     abc                  4        10
100     2020-10-19     abc                  3        2

结果中的输出权重应对应于最大dep_id。

~id ~from ~to ~label ~weight
100  A     B   knows    10

Answer 1

很难找到解决方案，因为您没有说明~id, ~from, ~to, ~label的计算方式。您应该可以使用window functions，即FIRST_VALUE()来实现所需的输出：

...
,CASE *some-condition* as "~label"
,FIRST_VALUE(weight)OVER(ORDER BY dep_id desc) "weight"
...

您可能需要添加PARTITION BY子句，这取决于您是希望整体获得第一个值，还是取决于其他一些条件。

聚合CASE语句中的列

1 个答案: