Question

如何计算2楼的Jane人数（不包括1楼的人）？

+------+---------+----------+
|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    |
+------+---------+----------+

预期结果为2，因为在2楼的Jane之前有2个人（Wally和Tommy）。

我尝试使用CHARINDEX从我生成的临时表中查找行号，但这似乎不起作用：

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

我认为一个简单的row_number()就可以解决问题

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

您可以这样做：

const result = [];
result.length = 2;

function combine(input, len, start) {
  if (len === 0) {
    console.log(result.join(" x "));
    return;
  }
  for (var i = start; i <= input.length - len; i++) {
    result[result.length - len] = input[i];
    combine(input, len - 1, i + 1);
  }
}

const Names = ['Name1', 'Name2', 'Name3', 'Name4', 'Name5'];

combine(Names, result.length, 0);

使用<div class="Generator"> <div id="button"></div> <button id="button1">Generate</button> </div>上的索引，我希望它比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);快。

计算满足条件时在列中位置的前几行

2 个答案: