mysql数据的操作

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LOAD DATA INFILE ‘文件路径和文件名‘ INTO TABLE 表名字;

或

source *.sql

　导出

SELECT 列1，列2 INTO OUTFILE ‘文件路径和文件名‘ FROM 表名字;

每个部门工资最高的员工

创建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 | Henry | 80000 | 2 |

| 3 | Sam | 60000 | 2 |

| 4 | Max | 90000 | 1 |

+----+-------+--------+--------------+

创建Department 表，包含公司所有部门的信息。

+----+----------+

| Id | Name |

+----+----------+

| 1 | IT |

| 2 | Sales |

+----+----------+

SELECT Department,Employee,Salary FROM
(SELECT d.Name as Department,e.Name as Employee,e.Salary FROM Employee e join Department d
on e.DepartmentId=d.Id) t
GROUP BY Department;

换座位

小美是一所中学的信息科技老师，她有一张 seat 座位表，平时用来储存学生名字和与他们相对应的座位 id。

其中纵列的 id 是连续递增的

小美想改变相邻俩学生的座位。

你能不能帮她写一个 SQL query 来输出小美想要的结果呢？

 请创建如下所示seat表：

示例：

+---------+---------+

| id | student |

+---------+---------+

| 1 | Abbot |

| 2 | Doris |

| 3 | Emerson |

| 4 | Green |

| 5 | Jeames |

+---------+---------+

假如数据输入的是上表，则输出结果如下：

+---------+---------+

| id | student |

+---------+---------+8

| 1 | Doris |

| 2 | Abbot |

| 3 | Green |

| 4 | Emerson |

| 5 | Jeames |

+---------+---------+

注意：

如果学生人数是奇数，则不需要改变最后一个同学的座位。

SELECT 
a.id,a.student
FROM
(SELECT id-1 AS id,student FROM seat WHERE id MOD 2=0
UNION
SELECT id,student FROM seat WHERE id MOD 2=1 and id =(SELECT COUNT(*) FROM seat)
UNION
SELECT id+1 as id,student FROM seat WHERE id MOD 2=1 and id!=(SELECT count(*) FROM seat)
)a
ORDER BY a.id

分数排名

编写一个 SQL 查询来实现分数排名。如果两个分数相同，则两个分数排名（Rank）相同。请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”。

创建以下score表：

+----+-------+

| Id | Score |

+----+-------+

| 1 | 3.50 |

| 2 | 3.65 |

| 3 | 4.00 |

| 4 | 3.85 |

| 5 | 4.00 |

| 6 | 3.65 |

+----+-------+

SELECT Score ,
(select count(distinct Score) from Score as s2 where s2.Score >= s1.Score) Rank 
FROM
score as s1
ORDER BY
Score 
DESC

行程和用户

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。

select t.Request_at as day,
round(sum(case when t.status=‘completed‘ then 0 else 1 end)/count(*),2) as ‘Cancellation Rate‘
from
 Trips t inner join users u1 on t.Client_Id=u1.Users_Id and u1.Banned=‘No‘
 where t.Request_at between ‘2013-10-01‘ and ‘2013-10-03‘
 group by t.Request_at;

各部门前3高工资的员工

+----+-------+--------+--------------+

| Id | Name | Salary | DepartmentId |

+----+-------+--------+--------------+

| 1 | Joe | 70000 | 1 |

| 2 | Henry | 80000 | 2 |

| 3 | Sam | 60000 | 2 |

| 4 | Max | 90000 | 1 |

| 5 | Janet | 69000 | 1 |

| 6 | Randy | 85000 | 1 |

+----+-------+--------+--------------+

编写一个 SQL 查询，找出每个部门工资前三高的员工。例如，根据上述给定的表格，查询结果应返回：

此外，请考虑实现各部门前N高工资的员工功能。

SELECT D1.Name Department, E1.Name Employee,  E1.Salary
FROM Employee E1, Employee E2, Department D1
WHERE E1.DepartmentID = E2.DepartmentID
AND E2.Salary >= E1.Salary 
AND E1.DepartmentID = D1.ID      
GROUP BY E1.Name
HAVING COUNT(DISTINCT E2.Salary) <= 3
ORDER BY D1.Name, E1.Salary DESC;

 分数排名 

+-------+------+

| Score | Rank |

+-------+------+

| 4.00 | 1 |

| 4.00 | 1 |

| 3.85 | 3 |

| 3.65 | 4 |

| 3.65 | 4 |

| 3.50 | 6 |

+-------+------

SELECT Score,
(SELECT count(*) FROM Score AS s2 WHERE s2.Score > s1.Score)+1 AS Rank 
FROM Score AS s1
ORDER BY Score DESC;

mysql数据的操作

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本文系统来源：https://www.cnblogs.com/MsHibiscus/p/10661312.html

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