Question

让我们说我知道我在SQL列中有一个特定的值，我想确定它的百分位数。基于另一个SO帖子（https://dba.stackexchange.com/questions/13703/get-the-rank-of-a-user-in-a-score-table）我尝试修改它，它可以工作但返回列中每个值的等级：

mysql> SELECT FIND_IN_SET( sum_squares, (SELECT GROUP_CONCAT(sum_squares ORDER BY sum_squares DESC)                  |
    -> FROM entries)                                                                                                 |
    -> ) AS rank                                                                                                     |
    -> FROM entries;

如下：

+------+                                                                                                             |
| rank |                                                                                                             |
+------+                                                                                                             |
|    4 |                                                                                                             |
|    4 |                                                                                                             |
|    2 |                                                                                                             |
|    3 |                                                                                                             |
|    1 |                                                                                                             |
+------+

我尝试通过在WHERE之前添加FROM entries语句来修改它，如下所示：

mysql> SELECT FIND_IN_SET( sum_squares, (SELECT GROUP_CONCAT(sum_squares ORDER BY sum_squares DESC)                  |
    -> FROM entries)                                                                                                 |
    -> ) AS rank          
    -> ) WHERE sum_squares = 5                                                                                             |
    -> FROM entries;

这会导致以下错误：

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE sum_squares = 5
FROM entries' at line 4

有人可以向我解释为什么我的查询似乎被禁止，如果还有其他方法可以做到这一点？我不想退回整个专栏！

这是一个示例entries表：

+----+-------------+
| id | sum_squares |
+----+-------------+
|  1 |           5 |
|  2 |           5 |
|  3 |          55 |
|  4 |          33 |
|  5 |         111 |
+----+-------------+

Answer 1

这是几乎在@ GordonLinoff的评论之后做了什么：

mysql> SELECT FIND_IN_SET( sum_squares, (SELECT GROUP_CONCAT( sum_squares ORDER BY sum_squares DESC)
    -> FROM entries HAVING sum_squares = 5)
    -> )  AS rank
    -> FROM entries

然而，这会返回匹配行中的答案以及其他行中的NULL，因此我欢迎对此进行改进。感谢。

+------+
| rank |
+------+
|    4 |
|    4 |
| NULL |
| NULL |
| NULL |
+------+

Answer 2

如果您想要单个条目的排名，请尝试以下查询：

SELECT COUNT(DISTINCT sum_squares) + 1
FROM entries
WHERE sum_squares > 5 ;

检索SQL列中特定值的百分位数

2 个答案: