中位数：也就是选取中间的数。一种衡量集中趋势的方法。
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要找中位数，首先需要从小到大排序，例如这组数据：23、29、20、32、23、21、33、25；
我们将数据排序20、21、23、23、25、29、32、33；排序后发现有8个数怎么办？
若有n个数，n为奇数，则选择第（n+1）/2个为中位数，若n为偶数，则中位数是（n/2以及n+1/2）的平均数
此例中选择24为中位数
有点复杂，在你基础上加了条有奇数的数据
创建表，插入数据：
create table test
(cat_id int,
price int);
insert into test values (101,90);
insert into test values (101,99);
insert into test values (102,98);
insert into test values (103,96);
insert into test values (102,95);
insert into test values (102,94);
insert into test values (102,93);
insert into test values (103,99);
insert into test values (103,98);
insert into test values (103,97);
insert into test values (104,96);
insert into test values (104,95);
insert into test values (105,97);
insert into test values (105,96);
insert into test values (105,95);
执行：
SELECT
t1.cat_id,
round(avg(t1.price), 1) price
FROM
(
SELECT
*
FROM
(
SELECT
t.cat_id,
t.price,
count(*) AS rank
FROM
test t
LEFT OUTER JOIN test r ON t.cat_id = r.cat_id
AND t.price = r.price
GROUP BY
t.cat_id,
t.price
ORDER BY
t.cat_id,
t.price DESC
) s
) t1,
(
SELECT DISTINCT
a.cat_id,
round(a.maxrank / 2) rank
FROM
(
SELECT
cat_id,
max(rank) maxrank,
MOD (max(rank), 2) modrank
FROM
(
SELECT
*
FROM
(
SELECT
t.cat_id,
t.price,
count(*) AS rank
FROM
test t
LEFT OUTER JOIN test r ON t.cat_id = r.cat_id
AND t.price = r.price
GROUP BY
t.cat_id,
t.price
ORDER BY
t.cat_id,
t.price DESC
) s
) t1
GROUP BY
cat_id
) a,
(
SELECT
*
FROM
(
SELECT
t.cat_id,
t.price,
count(*) AS rank
FROM
test t
LEFT OUTER JOIN test r ON t.cat_id = r.cat_id
AND t.price = r.price
GROUP BY
t.cat_id,
t.price
ORDER BY
t.cat_id,
t.price DESC
) s
) b
WHERE
a.cat_id = b.cat_id
AND a.modrank = 0
UNION ALL
SELECT DISTINCT
a.cat_id,
round(a.maxrank / 2) + 1 rank
FROM
(
SELECT
cat_id,
max(rank) maxrank,
MOD (max(rank), 2) modrank
FROM
(
SELECT
*
FROM
(
SELECT
t.cat_id,
t.price,
count(*) AS rank
FROM
test t
LEFT OUTER JOIN test r ON t.cat_id = r.cat_id
AND t.price = r.price
GROUP BY
t.cat_id,
t.price
ORDER BY
t.cat_id,
t.price DESC
) s
) t1
GROUP BY
cat_id
) a,
(
SELECT
*
FROM
(
SELECT
t.cat_id,
t.price,
count(*) AS rank
FROM
test t
LEFT OUTER JOIN test r ON t.cat_id = r.cat_id
AND t.price = r.price
GROUP BY
t.cat_id,
t.price
ORDER BY
t.cat_id,
t.price DESC
) s
) b
WHERE
a.cat_id = b.cat_id
AND a.modrank = 0
UNION ALL
SELECT DISTINCT
a.cat_id,
round(a.maxrank / 2) rank
FROM
(
SELECT
cat_id,
max(rank) maxrank,
MOD (max(rank), 2) modrank
FROM
(
SELECT
*
FROM
(
SELECT
t.cat_id,
t.price,
count(*) AS rank
FROM
test t
LEFT OUTER JOIN test r ON t.cat_id = r.cat_id
AND t.price = r.price
GROUP BY
t.cat_id,
t.price
ORDER BY
t.cat_id,
t.price DESC
) s
) t1
GROUP BY
cat_id
) a,
(
SELECT
*
FROM
(
SELECT
t.cat_id,
t.price,
count(*) AS rank
FROM
test t
LEFT OUTER JOIN test r ON t.cat_id = r.cat_id
AND t.price = r.price
GROUP BY
t.cat_id,
t.price
ORDER BY
t.cat_id,
t.price DESC
) s
) b
WHERE
a.cat_id = b.cat_id
AND a.modrank = 1
) t2
WHERE
t1.cat_id = t2.cat_id
AND t1.rank = t2.rank
GROUP BY
t1.cat_id
结果：
其中：
select * from (
select t.cat_id,t.price,count(*) as rank from test t
LEFT OUTER JOIN test r
on t.cat_id = r.cat_id
and t.price=r.price
group by t.cat_id,t.price
order by t.cat_id, t.price desc
) s
这条是主语句，主要是按照大小给出一个排名，然后根据中位数的公式，偶数的话，取最中间两个的平均数，奇数取最中间的数。自己研究一下吧。
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