在 SQL SERVER 中按组排名

Question

我需要按“String”和“Is_sep”对以下组进行排名，并按“Id”保持顺序： （SQL 服务器）

ID      | String    | Is_Sep    | Rank      | Rank_DESC
1       | XX        | 0         | 1         | 3
2       | XX        | 0         | 2         | 2
3       | XX        | 1         | 3         | 1
4       | XX        | 0         | 1         | 3
5       | XX        | 0         | 2         | 2
6       | XX        | 1         | 3         | 1
7       | YY        | 0         | 1         | 4
8       | YY        | 0         | 2         | 3
9       | YY        | 0         | 3         | 2
10      | YY        | 1         | 4         | 1
11      | ZZ        | 0         | 1         | 2
12      | ZZ        | 1         | 2         | 1
13      | ZZ        | 0         | 1         | 3
14      | ZZ        | 0         | 2         | 2
15      | ZZ        | 1         | 3         | 1

知道怎么做吗？

谢谢！

Answer 1

这是一种孤岛问题，is_sep 标志着每个孤岛的尽头。

我会使用窗口总和来定义组，然后row_number():

select 
    id,
    string,
    sep,
    row_number() over(partition by string, grp order by id) rn,
    row_number() over(partition by string, grp order by id desc) rn_desc
from (
    select 
        t.*,
        sum(is_sep) over(partition by string order by id desc) grp
    from mytable t
) t
order by id

请注意，对于此示例数据，您实际上并不需要分区中的string，因为is_sep 也表示string 值之间的转换。所以这同样适用：

select 
    id,
    string,
    is_sep,
    row_number() over(partition by grp order by id) rn,
    row_number() over(partition by grp order by id desc) rn_desc
from (
    select 
        t.*,
        sum(is_sep) over(order by id desc) grp
    from mytable t
) t
order by id

Demo on DB Fiddlde：

编号 |字符串 | is_sep | rn | rn_desc -: | :----- | -----: | -: | ------: 1 | XX | 0 | 1 | 3 2 | XX | 0 | 2 | 2 3 | XX | 1 | 3 | 1 4 | XX | 0 | 1 | 3 5 | XX | 0 | 2 | 2 6 | XX | 1 | 3 | 1 7 | YY | 0 | 1 | 4 8 | YY | 0 | 2 | 3 9 | YY | 0 | 3 | 2 10 | YY | 1 | 4 | 1 11 | ZZ | 0 | 1 | 2 12 | ZZ | 1 | 2 | 1 13 | ZZ | 0 | 1 | 3 14 | ZZ | 0 | 2 | 2 15 | ZZ | 1 | 3 | 1