Employee 表包含所有员工。Employee 表有三列：员工Id，公司名和薪水。
+-----+------------+--------+
|Id | Company | Salary |
+-----+------------+--------+
|1 | A | 2341 |
|2 | A | 341 |
|3 | A | 15 |
|4 | A | 15314 |
|5 | A | 451 |
|6 | A | 513 |
|7 | B | 15 |
|8 | B | 13 |
|9 | B | 1154 |
|10 | B | 1345 |
|11 | B | 1221 |
|12 | B | 234 |
|13 | C | 2345 |
|14 | C | 2645 |
|15 | C | 2645 |
|16 | C | 2652 |
|17 | C | 65 |
+-----+------------+--------+
请编写SQL查询来查找每个公司的薪水中位数。挑战点：你是否可以在不使用任何内置的SQL函数的情况下解决此问题。
+-----+------------+--------+
|Id | Company | Salary |
+-----+------------+--------+
|5 | A | 451 |
|6 | A | 513 |
|12 | B | 234 |
|9 | B | 1154 |
|14 | C | 2645 |
+-----+------------+--------+
题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/median-employee-salary
审题：求中位数，求一个单位工资在中间的人。
思考：可能有一个，奇数的时候；偶数的时候就会有两个中位数。不使用内置函数。
按照单位和工资排序，然后查找单位人数除以2的工资，并且提前判断奇偶数。
解题：
解法1：
薪水总数N是偶数时，下标(N-1)/2和N/2位置处为中位数。N是奇数时，下标(N-1)/2为中位数。
对每个公司group by分组，统计每组个数N，确定中位数区间[beg,beg+cnt]，beg从0开始。
beg = (N-1)/2
cnt = 0或1，N为偶数时为1，N为奇数时为0。
结果命名为表A：取到中位数的下标。COUNT(*)为每组总数，FLOOR向下取整函数。
(
SELECT E.Company,FLOOR((COUNT(*)-1)/2) AS `beg`,if(COUNT(*) % 2=1,0,1) AS `cnt`
FROM employee AS E
GROUP BY E.Company
) AS A
(select E.company,floor(count(*)-1)/2) as 'beg', if(count(*)%2 =1,0,1) as 'cnt') as A计算每个人薪水的升序排名。薪水比较方法：
if (A.salary = B.salary and A.id > B.id or A.salary > B.salary)
{
那么A的排名在B的排名后。
}员工表left join员工表，得出同一个公司中，排名在每个人之前的所有人:
SELECT *
FROM employee AS E1
LEFT JOIN employee AS E2
ON(E1.company = E2.company AND (E1.salary = E2.salary AND E1.Id > E2.Id OR E1.Salary > E2.Salary))按E1分组后，统计每组内，E2的个数即为E1的排名，排名从0开始。最后结果按薪水升序，命名为表B:
(
SELECT E1.Id,E1.Company,E1.Salary, COUNT(E2.Salary) AS `trank`
FROM employee AS E1
LEFT JOIN employee AS E2
ON(E1.company = E2.company AND (E1.salary = E2.salary AND E1.Id > E2.Id OR E1.Salary > E2.Salary))
GROUP BY E1.Id,E1.Company,E1.Salary
ORDER BY E1.Company,E1.Salary
) AS B
(
selcet e1.id,e1.company,e1,salary,count(e2.salary) as 'trank'
from employee as e1 left join employee as e2
on ()
) AS BSELECT B.Id,B.Company,B.Salary
FROM
(
SELECT E.Company,FLOOR((COUNT(*)-1)/2) AS `beg`,if(COUNT(*) % 2=1,0,1) AS `cnt`
FROM employee AS E
GROUP BY E.Company
) AS A
JOIN (
SELECT E1.Id,E1.Company,E1.Salary, COUNT(E2.Salary) AS `trank`
FROM employee AS E1
LEFT JOIN employee AS E2
ON(E1.company = E2.company AND (E1.salary = E2.salary AND E1.Id > E2.Id OR E1.Salary > E2.Salary))
GROUP BY E1.Id,E1.Company,E1.Salary
ORDER BY E1.Company,E1.Salary
) AS B
ON (A.company = B.company AND B.trank BETWEEN A.beg AND (A.beg+A.cnt))解法2：
思路与解法一相同。换了一种计算公司内部人员排名的方法。
解法一采用表left join，分组再计数的方法。
这里，采用用户变量的方法，简化排名的计算。
定义用户变量：pre_company——上一行的公司字段。trank——当前行的排名，从0开始。
现在有了中位数区间和每个人的排名。
连接表A和表B，公司相同，B.trank在[A.beg,A.beg+A.cnt]之间的行，为中位数：
(SELECT @pre_company:= NULL,@trank:=0) AS T计算排名的逻辑：
if (当前行的company = pre_company){
trank = trank + 1
}else{
trank = 0
}SQL代码，结果命名为表A：本题的难点。
(
SELECT
id,
company,
salary,
@trank:=IF(@pre_company=company,@trank+1,0) AS `trank`,
@pre_company:=company
FROM employee,(SELECT @pre_company:= NULL,@trank:=0) AS T
ORDER BY company,salary
) AS A计算中位数下标的方法如解法一。
薪水总数N是偶数时，下标(N-1)/2和N/2位置处为中位数。N是奇数时，下标(N-1)/2为中位数。
对每个公司group by分组，统计每组个数N，确定中位数区间[beg,beg+cnt]，beg从0开始。
beg = (N-1)/2
cnt = 0或1，N为偶数时为1，N为奇数时为0。
结果命名为表B：
(
SELECT company, FLOOR((COUNT(*)-1)/2) AS `beg`, if(COUNT(*)%2=1,0,1) AS `cnt`
FROM employee
GROUP BY company
) AS B连接表A和表B ，公司相同，B.trank在[A.beg,A.beg+A.cnt]之间的行，为中位数：
SELECT A.Id,A.Company,A.Salary
FROM
(
SELECT
id,
company,
salary,
@trank:=IF(@pre_company=company,@trank+1,0) AS `trank`,
@pre_company:=company
FROM employee,(SELECT @pre_company:= NULL,@trank:=0) AS T
ORDER BY company,salary
) AS A
JOIN
(
SELECT company, FLOOR((COUNT(*)-1)/2) AS `beg`, if(COUNT(*)%2=1,0,1) AS `cnt`
FROM employee
GROUP BY company
) AS B
ON (A.company=B.company AND A.trank BETWEEN B.beg AND B.beg+B.cnt)知识点：
floor：函数只返回整数部分，小数部分舍弃。
round：函数四舍五入，大于0.5的部分进位，不到则舍弃。
BETWEEN 运算符用于 WHERE 表达式中，选取介于两个值之间的数据范围。BETWEEN 同 AND 一起搭配使用。
