SQL:如何从重复行中选择第一条记录？

Question

在执行以下查询以查找重复项时

select * from (
select a.* ,count (*) over (partition by a.ID) as tot
from HREMP a 
) tt
where tt.tot >1

它返回 423 行，

我执行了另一个查询来查找不重复的记录

  select * from (
select a.* ,count (*) over (partition by a.ID) as tot
from HREMP a 
) tt
where tt.tot =1

返回 685 条记录

I found that there are 196 distinct records among the 423 duplicateNow, How to select the first record from duplicate records?

Answer 1

select distinct * 
from ( select a.*, count(*) over (partition by a.ID) as tot
       from HREMP a 
     ) tt
where tt.tot > 1

或

select * 
from ( select a.*
            , count(*)     over (partition by a.ID) as tot
            , row_number() over (partition by a.ID order by 1) as rn
       from HREMP a 
     ) tt
where tt.tot > 1 
and   tt.rn = 1