1. 排序

MySQL内置函数：（直接获取排名）
例如成绩：{99，99，85，80，75}

  • DENSE_RANK()。如果使用 DENSE_RANK() 进行排名会得到：1，1，2，3，4。
  • RANK()。如果使用 RANK() 进行排名会得到：1，1，3，4，5。
  • ROW_NUMBER()。如果使用 ROW_NUMBER() 进行排名会得到：1，2，3，4，5。
    1. select score, DENSE_RANK() OVER (ORDER BY Score DESC) as 'Rank'
    2. from Scores;
    连接查询：（转换排名为计数）
    要求：分数相同排名相同,排名不间断（如1, 2, 2, 3...）
输入：
+----+-------+ 
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+  
输出
+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+
解答：
select S1.Score, count(distinct(S2.Score)) AS `Rank` 
from Scores AS S1 join Scores AS S2 on S1.score <= S2.Score
group by S1.Id
order by `Rank` ASC

    例题

    （1）前几名

    ```sql 题目： Employee 表包含所有员工信息， 每个员工有其对应的工号 Id，姓名 Name，工资 Salary 和部门编号 DepartmentId 。

+——+———-+————+———————+ | Id | Name | Salary | DepartmentId | +——+———-+————+———————+ | 1 | Joe | 85000 | 1 | | 2 | Henry | 80000 | 2 | | 3 | Sam | 60000 | 2 | | 4 | Max | 90000 | 1 | | 5 | Janet | 69000 | 1 | | 6 | Randy | 85000 | 1 | | 7 | Will | 70000 | 1 | +——+———-+————+———————+ Department 表包含公司所有部门的信息。

+——+—————+ | Id | Name | +——+—————+ | 1 | IT | | 2 | Sales | +——+—————+ 编写一个 SQL 查询，找出每个部门获得前三高工资的所有员工。 例如，根据上述给定的表，查询结果应返回：

+——————+—————+————+ | Department | Employee | Salary | +——————+—————+————+ | IT | Max | 90000 | | IT | Randy | 85000 | | IT | Joe | 85000 | | IT | Will | 70000 | | Sales | Henry | 80000 | | Sales | Sam | 60000 | +——————+—————+————+

```sql
/**
法一解题思路:
转换排名为计数，
外查询，高于工资值的数量应<3 
子查询，查询同一部门比外查询工资值高的数量
**/
SELECT
    d.Name AS 'Department', e1.Name AS 'Employee', e1.Salary
FROM
    Employee e1
        JOIN
    Department d ON e1.DepartmentId = d.Id
WHERE
    3 > (SELECT
            COUNT(DISTINCT e2.Salary)
        FROM
            Employee e2
        WHERE
            e2.Salary > e1.Salary
                AND e1.DepartmentId = e2.DepartmentId
        )


/**
法二解题思路:
先对Employee表进行部门分组工资排名，
再关联Department表查询部门名称，
再使用WHERE筛选出排名小于等于3的数据（也就是每个部门排名前3的工资）。
**/
SELECT 
B.Name AS Department,
A.Name AS Employee,
A.Salary
FROM (SELECT DENSE_RANK() OVER (partition by DepartmentId order by Salary desc) 
            AS ranking,DepartmentId,Name,Salary
      FROM Employee) AS A
JOIN Department AS B ON A.DepartmentId=B.id
WHERE A.ranking<=3

（2）第N高

使用 order + limit + offset

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  set N := N-1;
  RETURN (
      # Write your MySQL query statement below.
      SELECT
        (SELECT DISTINCT
                Salary
            FROM
                Employee
            ORDER BY Salary DESC
            LIMIT 1 OFFSET N)
      AS SecondHighestSalary
  );
END

2.连接查询

两种方式：

  • select XX from A,B where A.x = B.x
  • select XX from A left join B on A.x = B.x

例一：

题目：
编写一个 SQL 查询，查找所有至少连续出现三次的数字。

+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+
例如，给定上面的 Logs 表， 1 是唯一连续出现至少三次的数字。

+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+

/* 解答 */
SELECT DISTINCT l1.Num as ConsecutiveNums
FROM
    Logs l1,
    Logs l2,
    Logs l3
WHERE
    l1.Id = l2.Id - 1
    AND l2.Id = l3.Id - 1
    AND l1.Num = l2.Num
    AND l2.Num = l3.Num

3. 复杂查询

多值IN

题目：
Employee 表包含所有员工信息。
每个员工有其对应的 Id, salary 和 department Id。

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+
Department 表包含公司所有部门的信息。

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+
编写一个 SQL 查询，找出每个部门工资最高的员工。
对于上述表，您的 SQL 查询应返回以下行（行的顺序无关紧要）。

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
+------------+----------+--------+

/**
    嵌套查询，
  子查询中查出来各个部门最高的工资，
  外查询:IN要同时满足（部门和工资）
**/
SELECT
    Department.name AS 'Department',
    Employee.name AS 'Employee',
    Salary
FROM
    Employee
        JOIN
    Department ON Employee.DepartmentId = Department.Id
WHERE
    (Employee.DepartmentId , Salary) IN
    (   SELECT
            DepartmentId, MAX(Salary)
        FROM
            Employee
        GROUP BY DepartmentId
    )

IF语句

题目：
Trips 表中存所有出租车的行程信息。每段行程有唯一键 Id，Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外键。Status 是枚举类型，枚举成员为 (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’)。

+----+-----------+-----------+---------+--------------------+----------+
| Id | Client_Id | Driver_Id | City_Id |        Status      |Request_at|
+----+-----------+-----------+---------+--------------------+----------+
| 1  |     1     |    10     |    1    |     completed      |2013-10-01|
| 2  |     2     |    11     |    1    | cancelled_by_driver|2013-10-01|
| 3  |     3     |    12     |    6    |     completed      |2013-10-01|
| 4  |     4     |    13     |    6    | cancelled_by_client|2013-10-01|
| 5  |     1     |    10     |    1    |     completed      |2013-10-02|
| 6  |     2     |    11     |    6    |     completed      |2013-10-02|
| 7  |     3     |    12     |    6    |     completed      |2013-10-02|
| 8  |     2     |    12     |    12   |     completed      |2013-10-03|
| 9  |     3     |    10     |    12   |     completed      |2013-10-03| 
| 10 |     4     |    13     |    12   | cancelled_by_driver|2013-10-03|
+----+-----------+-----------+---------+--------------------+----------+
Users 表存所有用户。每个用户有唯一键 Users_Id。Banned 表示这个用户是否被禁止，Role 则是一个表示（‘client’, ‘driver’, ‘partner’）的枚举类型。

+----------+--------+--------+
| Users_Id | Banned |  Role  |
+----------+--------+--------+
|    1     |   No   | client |
|    2     |   Yes  | client |
|    3     |   No   | client |
|    4     |   No   | client |
|    10    |   No   | driver |
|    11    |   No   | driver |
|    12    |   No   | driver |
|    13    |   No   | driver |
+----------+--------+--------+
写一段 SQL 语句查出 2013年10月1日 至 2013年10月3日 期间非禁止用户的取消率。基于上表，你的 SQL 语句应返回如下结果，取消率（Cancellation Rate）保留两位小数。

取消率的计算方式如下：(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)

+------------+-------------------+
|     Day    | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 |       0.33        |
| 2013-10-02 |       0.00        |
| 2013-10-03 |       0.50        |
+------------+-------------------+

            / 
            COUNT(T.STATUS),
            2
    ) AS `Cancellation Rate`
FROM Trips AS T
JOIN Users AS U1 ON (T.client_id = U1.users_id AND U1.banned ='No')
JOIN Users AS U2 ON (T.driver_id = U2.users_id AND U2.banned ='No')
WHERE T.request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY T.request_at