平均数、中位数、众数常用来描述数据的集中程度，在mysql中，均值的计算较为简单，但中位数和众数尚不支持可以直接计算的函数，在这里向大家介绍一下中位数和众数的计算方法。

本文所使用的数据表为score，包含三个字段：
s_id 为学生id
c_id 为课程id
s_score 为课程成绩

1、中位数
查找课程编号为‘02’号的学生成绩中位数
课程编号为‘02’的课程成绩为

mysql> select s_score from score where c_id='02';
+---------+
| s_score |
+---------+
|      90 |
|      60 |
|      80 |
|      30 |
|      87 |
|      89 |
+---------+
6 rows in set (0.00 sec)

查找中位数

mysql> select avg(c.s_score) from(
    ->   select a.s_score from score a,score b
    ->   where a.c_id=b.c_id and a.c_id='02'
    ->   group by a.s_score
    ->   having sum(case when a.s_score=b.s_score then 1 else 0 end)
    ->          >= abs(sum(sign(a.s_score-b.s_score)))
    ->   )c;
+----------------+
| avg(c.s_score) |
+----------------+
|        83.5000 |
+----------------+
1 row in set (0.00 sec)

解析：
当把having条件作为查询内容时，便比较清晰了，结果如下

mysql> select a.s_score,
    -> sum(case when a.s_score=b.s_score then 1 else 0 end )as if_equal,
    -> abs(sum(sign(a.s_score-b.s_score)))
    -> from score a,score b where a.c_id=b.c_id and a.c_id='02'
    -> group by a.s_score;
+---------+----------+-------------------------------------+
| s_score | if_equal | abs(sum(sign(a.s_score-b.s_score))) |
+---------+----------+-------------------------------------+
|      30 |        1 |                                   5 |
|      60 |        1 |                                   3 |
|      80 |        1 |                                   1 |
|      87 |        1 |                                   1 |
|      89 |        1 |                                   3 |
|      90 |        1 |                                   5 |
+---------+----------+-------------------------------------+
6 rows in set (0.00 sec)

2、众数
一组数据可以存在多个众数

mysql> select s_score from score where c_id='02';
+---------+
| s_score |
+---------+
|      90 |
|      60 |
|      80 |
|      30 |
|      80 |
|      89 |
+---------+
6 rows in set (0.00 sec)

mysql> select s_score from score
    -> where c_id='02'
    -> group by s_score
    -> having count(*)
    ->        >= all(select count(*) from score where c_id='02' group by s_score);
+---------+
| s_score |
+---------+
|      80 |
+---------+
1 row in set (0.00 sec)

解析：

mysql> select s_score,count(*) from score where c_id='02' group by s_score;
+---------+----------+
| s_score | count(*) |
+---------+----------+
|      30 |        1 |
|      60 |        1 |
|      80 |        2 |
|      89 |        1 |
|      90 |        1 |
+---------+----------+
5 rows in set (0.00 sec)

count() >= all(select count() from score where c_id=’02’ group by s_score)
表示score中分组后某个s_score出现的次数 >= 所有不同s_score出现的次数。