数据导入导出
导入table
http://www.runoob.com/mysql/mysql-database-import.html导出table
http://www.runoob.com/mysql/mysql-database-export.html
SELECT * FROM runoob_tbl
INTO OUTFILE '/tmp/runoob.txt'设置分隔符
SELECT a,b,a+b INTO OUTFILE '/tmp/result.text'
FIELDS TERMINATED BY ',' OPTIONALLY ENCLOSED BY '"'
LINES TERMINATED BY '\n'
FROM test_table;作业
- 1.项目七: 各部门工资最高的员工（难度：中等）
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 |
+----+----------+编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
解答：
CREATE TABLE employee(
Id INT NOT NULL PRIMARY KEY,
Name VARCHAR(20) NOT NULL,
Salary INT,
DepartmentId INT
);
INSERT INTO employee
(Id,Name,Salary,DepartmentId)
VALUES
(1,'Joe',70000,1),
(2,'Henry',80000,2),
(3,'Sam',60000,2),
(4,'Max',90000,2);
CREATE TABLE Department(
Id INT NOT NULL,
Name VARCHAR(20)
);
INSERT INTO Department
(Id,Name)
VALUES
(1,'IT'),
(2,'Sales');
SELECT Department.Name AS Department, employee.Name As Employee, employee.Salary AS Salary
FROM Department
JOIN employee
WHERE Department.Id = employee.DepartmentId
GROUP BY employee.DepartmentId
HAVING MAX(Salary)结果：
- 2.项目八: 换座位（难度：中等）
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 |
+---------+---------+
| 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。
解答：
CREATE TABLE seat(
id INT NOT NULL PRIMARY KEY,
student VARCHAR(20)
);
INSERT INTO seat
VALUES
(1,'Abbot'),
(2,'Doris'),
(3,'Emerson'),
(4,'Green'),
(5,'Jeames');
SELECT s.id,s.student from (
SELECT id-1 as id, student
FROM seat
WHERE id%2 = 0
UNION
SELECT id+1 as id, student
FROM seat
WHERE id%2 =1 AND id != (SELECT MAX(id) FROM seat)
UNION
SELECT id,student
FROM seat
WHERE id%2=1 AND id = (SELECT MAX(id) FROM seat);
)s
ORDER BY s.id;结果：
- 3.项目九: 分数排名（难度：中等）
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 |
+----+-------+例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：
+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
+-------+------+CREATE TABLE score(
id int,
score DECIMAL(3,2)
);
INSERT INTO score
VALUES
(1,3.50),
(2,3.65),
(3,4.00),
(4,3.85),
(5,4.00),
(6,3.65);
SELECT id, score,
CASE
WHEN @pre_score = score THEN @rank_tmp
WHEN @pre_score:=score THEN @rank_tmp:=@rank_tmp+1
END AS 'rank'
FROM score,(SELECT @rank_tmp:=0,@pre_score:=NULL) tmp
ORDER BY
score.score DESC;
结果：
这里用了变量和 CASE WHEN 语句，十分巧妙。
在MySQL 8.x 版本引入了窗口函数，包括 rank()、dense_rank()；所以可以这样写
SELECT
id,
score,
dense_rank ( ) over ( ORDER BY score.score DESC ) AS 'rank'
FROM
score
ORDER BY
score DESC;轻松优雅美滋滋。
参考
- 排序部分参考了 https://www.jianshu.com/p/bb1b72a1623e
- 换座位参考了 https://blog.csdn.net/wal1314520/article/details/80117658
- 窗口函数参考了 https://blog.51cto.com/11103985/2341971