在论坛中出现的比较难的sql问题：4(row_number函数+子查询 分组连续编号问题)

所以，觉得有必要记录下来，这样以后再次碰到这类问题，也能从中获取解答的思路。

求一查询语句 

 http://bbs.csdn.net/topics/390633004

CREATE #temp (cName CHAR(1),re int)
INSERT #temp
SELECT ‘A‘,1 UNION ALL
SELECT ‘A‘,2 UNION ALL
SELECT ‘B‘,3 UNION ALL
SELECT ‘A‘,4 UNION ALL
SELECT ‘A‘,5

如何查询得到如下的结果：
cName,re,xh
-------------------
‘A‘,1,1
‘A‘,2,1
‘B‘,3,2
‘A‘,4,3
‘A‘,5,3

下面是我的解答：

select *,
       dense_rank() over(order by case when exists(select  t2.re from #temp t2 
                                                   where t1.cname = t2.cname 
                                                   and t1.re= t2.re + 1)
                                            then (select  t2.re from #temp t2 
                                                   where t1.cname = t2.cname 
                                                   and t1.re= t2.re + 1)
                                       else t1.re
                                   end
                         ) as xh
from #temp t1
/*
cName re xh
A	  1	 1
A	  2	 1
B	  3	 2
A	  4	 3
A	  5	 3
*/

但是这个解答是有问题的，因为当连续的记录超过3条时，就会有问题，

所以修改了一下，这个是正确的解法：

create table #temp (cName CHAR(1),re int)
 
insert into #temp
SELECT ‘A‘,1 UNION ALL
SELECT ‘A‘,2 UNION ALL
SELECT ‘B‘,3 UNION ALL
SELECT ‘A‘,4 UNION ALL
SELECT ‘A‘,5 union all
SELECT ‘A‘,6 union all
SELECT ‘A‘,7 union all
SELECT ‘D‘,8 union all
SELECT ‘D‘,9 union all
SELECT ‘D‘,10 union all
select ‘B‘,11 union all
select ‘A‘,12
 
;with t
as
(
select *,
       row_number() over(partition by cname order by re) as rownum
from #temp 
)
 
select cname,
       re,
       dense_rank() over(order by case when exists(select min(t2.re) from t t2 
                                                   where t1.cname = t2.cname 
                                                   and t1.re-t1.rownum= t2.re-t2.rownum)
                                            then (select  min(t2.re) from t t2 
                                                   where t1.cname = t2.cname 
                                                   and t1.re-t1.rownum= t2.re-t2.rownum)
                                       else t1.re
                                   end
                         ) as xh
from t t1
/*cname	re	xh
A	1	1
A	2	1
B	3	2
A	4	3
A	5	3
A	6	3
A	7	3
D	8	4
D	9	4
D	10	4
B	11	5
A	12	6
*/

在论坛中出现的比较难的sql问题：4(row_number函数+子查询 分组连续编号问题)

原文：https://www.cnblogs.com/lonelyxmas/p/12019979.html