Question

鉴于以下table_a：

 id  | status | reg_date | so_date
 ----+------------------------------
  1  |   abc  | 15-11-01 | 15-11-02 
  1  |   def  | 15-11-01 | 15-11-04
  1  |   abc  | 15-11-01 | 15-11-06
  2  |   abc  | 15-11-01 | 15-11-03
  2  |   abc  | 15-11-01 | 15-11-04
  2  |   abc  | 15-11-01 | 15-11-06
  2  |   abc  | 15-11-01 | 15-11-08
  3  |   xyz  | 15-11-01 | 15-11-03
  3  |   def  | 15-11-01 | 15-11-08
  3  |   def  | 15-11-01 | 15-11-09

以下是必需的：对于每个id组，so_date和reg_date之间的最小差异，如果status是＆＃34; abc＆＃34;。这产生了下表：

 id  | min_date  
 ----+----------
  1  |    1   
  2  |    2   
  3  |   Null

以下代码就足够了：

select 
    id,
    distinct datediff('day', reg_date, min(so_date) over (partition by id)) as min_date
from table_a  
where status ilike '%abc%'

但是，由于其他＆＃34;选择＆＃34;的限制，在基表使用过滤器where status ilike '%abc%'是不可行的。我尝试使用

 select 
        id,
        case when status is like '%abc%' then (distinct datediff('day', reg_date, min(so_date) over (partition by id))) end as min_date
    from table_a

但是这没用。任何见解/建议将不胜感激。感谢

Answer 1

根据你的样本应

select 
    id
  , min (datediff( so_date, reg_date )) as min_date
from table_a 
where status ='abc'
group by id

如果你不想要蝙蝠案件那么

select 
   case status when 'abc' then  id end
  , case status when 'abc' then min (datediff( so_date, reg_date ))  end as min_date
from table_a 
group by id

在sql中使用带PARTITION的CASE语句

1 个答案: