Question

select distinct a.pattern ,convert(numeric(18,0) ,isnull(b.[DCount]as [DCount]
,DENSE_RANK () OVER ( Partition by a.pattern order by b.[DCount]desc ) as [Rank]  
from a, b

将输出显示为

Pattern Dcount  Rank
A   0   1
A   0   1
A   0   2
A   0   2
B   0   2
B   0   2
B   0   2
B   0   2
C   0   2
C   0   2
C   0   2

所需输出为

Pattern Dcount  Rank
A   0   1
A   0   1
A   0   1
A   0   1
B   0   1
B   0   1
B   0   1
B   0   1
C   0   1
C   0   1
C   0   1

Answer 1

以下面的方式使用DENSE_RANK

select * ,DENSE_RANK () OVER ( order by Pattern, Dcount) as rn from t

Answer 2

也许您只想按dcount进行排名。如果是这样：

select distinct a.pattern, convert(numeric(18,0),
       coalesce(b.DCount, 0) as DCount,
       dense_rank() over (order by b.[DCount] desc) as [Rank]  
from a cross join b;

Answer 3

尝试一下：您需要按模式，dcount进行分区，并仅按模式进行排序

demo

select *,dense_Rank() over(partition by Pattern,dcount order by Pattern) from tablename

等级函数给出错误的等级

3 个答案: