計算滿足條件時列中position前面有多少行

Question

如何計算 2 樓的 Jane 前面有多少人（不包括 1 樓的人）？

+------+---------+----------+
|Index |  Name   |   Floor  |
+------+---------+----------+
|   1  | Sally   |     1    |
|   2  | Sue     |     1    |
|   3  | Fred    |     1    |
|   4  | Wally   |     2    |
|   5  | Tommy   |     2    |
|   6  | Jane    |     2    |
|   7  | Bart    |     2    |
|   8  | Sam     |     3    |
+------+---------+----------+

預期結果是 2，因為在 2 樓 Jane 前面有 2 個人（Wally 和 Tommy）。

我嘗試使用 CHARINDEX 從我生成的臨時表中查找行號，但這似乎不起作用：

SELECT CHARINDEX('Jane', Name) as position
INTO #test
FROM tblExample
WHERE Floor = 2
select ROW_NUMBER() over (order by position) from #test
WHERE position = 1

Answer 1

我認為一個簡單的row_number()就可以了

Select Value = RN-1
 From  (
         Select *
               ,RN = row_number() over (partition by [floor] order by [index])
         From  YourTable
         Where [Floor]=2
       ) A
 Where [Name]='Jane'

Answer 2

你可以這樣做：

select count(*)
from t
where t.floor = 2 and
      t.id < (select t2.id from t t2 where t2.name = 'Jane' and t2.floor = 2);

在(floor, name, id)上有一個索引，我希望這比row_number()快。