Question

需要sql select min（date），max（date）更改每个代码的stat 我尝试这个查询它适用于哪里条件但他给我错误的数据没有条件

select x.code,x.stat,x.grp, min(date) as min_date, max(date) as max_date
from (
    select code,stat,date ,
          row_number() over (order by date)- row_number() over (partition by stat order by date) as grp
    from  c 
     where code='1')  X
   group by code,stat,grp
   order by min(date)

我的桌子

code     stat     date
1         s       01/01/2012
1         s       02/01/2012
1         i       03/01/2012
1         s       04/01/2012
2         a       01/01/2012
2         a       02/01/2012

需要此结果

code     stat     min date     max date
1         s       01/01/2012  02/01/2012
1         i       03/01/2012  03/01/2012
1         s       04/01/2012  04/01/2012
2         a       01/01/2012  02/01/2012

Answer 1

我认为您的代码基本上是正确的，除了缺少的partition by：

select c.code, c.stat, c.grp,
       min(c.date) as min_date, max(c.date) as max_date
from (select c.*
             row_number() over (partition by c.code order by c.date) as seqnum,
             row_number() over (partition by c.code, c.stat order by date) as seqnum_d
      from c 
     ) c
group by c.code, c.stat, c.grp
order by min(c.date);

SQL查询在数据更改时查找max，min

1 个答案: