首先公布一下DAY4的作业答案。

项目五：

--创建表create table person (personid int not null primary key,firstname varchar(20) not null,lastname varchar(20) not null);--插入数据insert into person values (101, 'ming', 'Yao');insert into person values (102, 'lei', 'Wu');insert into person values (103, 'zhi', 'Zheng');--查询结果mysql> select * from person;+----------+-----------+----------+| personid | firstname | lastname |+----------+-----------+----------+|      101 | ming      | Yao      ||      102 | lei       | Wu       ||      103 | zhi       | Zheng    |+----------+-----------+----------+3 rows in set (0.00 sec)--创建表create table address (addressid int not null primary key,personid int not null,city varchar(20) not null,state varchar(20) not null);--插入数据insert into address values (8000, 9999, 'xian', 'UA');insert into address values (8008, 102, 'beijing', 'AB');insert into address values (9009, 103, 'hangzhou', 'AC');--查询结果mysql> select * from address;+-----------+----------+----------+-------+| addressid | personid | city     | state |+-----------+----------+----------+-------+|      8000 |     9999 | xian     | UA    ||      8008 |      102 | beijing  | AB    ||      9009 |      103 | hangzhou | AC    |+-----------+----------+----------+-------+3 rows in set (0.00 sec)mysql> select firstname, lastname, a.city, a.state    -> from person    -> left join address as a on person.personid = a.personid;+-----------+----------+----------+-------+| firstname | lastname | city     | state |+-----------+----------+----------+-------+| ming      | Yao      | NULL     | NULL  || lei       | Wu       | beijing  | AB    || zhi       | Zheng    | hangzhou | AC    |+-----------+----------+----------+-------+3 rows in set (0.00 sec)

项目六：

#问题分析a. 删除重复的电子邮箱b. 需要保留ID最小的那个（删除的ID比留下的那一条ID要大）c. 考察自连接知识，把一张表当两张表来用。--code--mysql> select * from email;+----+---------+| ID | Email   |+----+---------+|  1 | a@b.com ||  2 | c@d.com ||  3 | a@b.com |+----+---------+3 rows in set (0.00 sec)mysql> delete e1    -> from email as e1, email as e2    -> where e1.Email = e2.Email    -> and e1.ID > e2.ID;Query OK, 1 row affected (0.01 sec)mysql> select * from email;+----+---------+| ID | Email   |+----+---------+|  1 | a@b.com ||  2 | c@d.com |+----+---------+2 rows in set (0.00 sec)

继续今日DAY5内容：

1. 数据导入导出（公众号回复em获取文件）

• navicat导入导出方法：
  连接数据库账户
  选择数据world
  导入向导
  导入类型选择excel
  导入过程中可以修改字段类型
  导入结果：

• 导出步骤：

选择数据库

选择表

导出向导

导出结果：

2. 作业

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  |+------------+----------+--------+

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。

答案明天揭晓。

喜欢就给个好看吧~~