MySQL基础练习题：创建数据库

这部分主要是为了帮助大家回忆回忆MySQL的基本语法，数据库来自于MySQL的官方简化版，题目也是网上非常流行的35题。这些基础习题基本可以涵盖面试中需要现场写SQL的问题。

创建数据库

在开始练习之前，我默认你的电脑上是没有本系列练习题需要的数据库。

第一步，准备数据

我这里写好一个.sql文件，放在了网盘里面

链接：https://pan.baidu.com/s/19r4ap0RBT95ghSN9VSui2w?pwd=aigo
提取码：aigo

大小仅仅2KB，下载应该会很快。

我将这个文件放在了 /usr/src/ 下面

[09:16:13 root@MySQL-server src]#ls
data.sql  debug  kernels
[09:16:15 root@MySQL-server src]#

那么该文件要如何使用呢？

第二步，创建数据库

接下来登陆MySQL，

(root@localhost) [(none)]>show databases;
+--------------------+
| Database           |
+--------------------+
| book               |
| information_schema |
| mysql              |
| performance_schema |
| springbootdb       |
| sys                |
+--------------------+
6 rows in set (0.03 sec)

(root@localhost) [(none)]>

那么我需要建一个数据库，就叫employees，一系列命令如下。

(root@localhost) [(none)]>create database employees;
Query OK, 1 row affected (0.10 sec)

(root@localhost) [(none)]>use employees;
Database changed
(root@localhost) [employees]>show tables;
Empty set (0.14 sec)


(root@localhost) [employees]>

第三步，导入数据

很简单且朴实无华的一条命令

source /usr/src/data.sql

执行过后

(root@localhost) [employees]>show tables;
+---------------------+
| Tables_in_employees |
+---------------------+
| dept                |
| emp                 |
| salgrade            |
+---------------------+
3 rows in set (0.00 sec)

该数据库中就出现了三张表

那么分别对三张表执行select *

(root@localhost) [employees]>select * from dept;
+--------+-------------+----------+
| deptno | dname       | loc      |
+--------+-------------+----------+
|     10 | ACCOUNTING  | NEW YORK |
|     20 | RESEARCHING | DALLAS   |
|     30 | SALES       | CHICAGO  |
|     40 | OPERATIONS  | BOSTON   |
+--------+-------------+----------+
4 rows in set (0.03 sec)

(root@localhost) [employees]>

(root@localhost) [employees]>select * from salgrade;
+-------+-------+-------+
| grade | losal | hisal |
+-------+-------+-------+
|     1 |   700 |  1200 |
|     2 |  1201 |  1400 |
|     3 |  1401 |  2000 |
|     4 |  2001 |  3000 |
|     5 |  3001 |  5000 |
+-------+-------+-------+
5 rows in set (0.05 sec)

(root@localhost) [employees]>


(root@localhost) [employees]>select * from emp;
+-------+--------+-----------+------+------------+---------+---------+--------+
| empno | ename  | job       | mgr  | hiredate   | sal     | comm    | deptno |
+-------+--------+-----------+------+------------+---------+---------+--------+
|  7369 | SIMITH | CLERK     | 7902 | 1980-12-17 |  800.00 |    NULL |     20 |
|  7499 | ALLEN  | SALESMAN  | 7698 | 1981-02-20 | 1600.00 |  300.00 |     30 |
|  7521 | WARD   | SALESMAN  | 7698 | 1981-02-22 | 1250.00 |  500.00 |     30 |
|  7566 | JONES  | MANAGER   | 7839 | 1981-04-02 | 2975.00 |    NULL |     20 |
|  7654 | MARTIN | SALESMAN  | 7698 | 1981-09-28 | 1250.00 | 1400.00 |     30 |
|  7698 | BLAKE  | MANAGER   | 7839 | 1981-05-01 | 2850.00 |    NULL |     30 |
|  7782 | CLARK  | MANAGER   | 7839 | 1981-06-09 | 2450.00 |    NULL |     10 |
|  7788 | SCOTT  | ANALYST   | 7566 | 1987-04-19 | 3000.00 |    NULL |     20 |
|  7839 | KING   | PRESIDENT | NULL | 1981-11-17 | 5000.00 |    NULL |     10 |
|  7844 | TURNER | SALESMAN  | 7698 | 1981-09-08 | 1500.00 |    NULL |     30 |
|  7876 | ADAMS  | CLERK     | 7788 | 1987-05-23 | 1100.00 |    NULL |     20 |
|  7900 | JAMES  | CLERK     | 7698 | 1981-12-03 |  950.00 |    NULL |     30 |
|  7902 | FORD   | ANALYST   | 7566 | 1981-12-03 | 3000.00 |    NULL |     20 |
|  7934 | MILLER | CLERK     | 7782 | 1982-01-23 | 1300.00 |    NULL |     10 |
+-------+--------+-----------+------+------------+---------+---------+--------+
14 rows in set (0.00 sec)

(root@localhost) [employees]>

简单练习一下

那今天就先做一道简单的练习题

取得每个部门最高薪水的人员名称

首先分析一下题目，该题目需要返回的是人员名称，其次该人员是该部门薪资最高的人。人员名称，部门以及薪资都在emp一个表里，那么最多也就是自己join自己

拆解

第一步，我就去看看每个部门最高的薪水。

(root@localhost) [employees]>select deptno,max(sal) as maxsal from emp group by deptno;
+--------+---------+
| deptno | maxsal  |
+--------+---------+
|     20 | 3000.00 |
|     30 | 2850.00 |
|     10 | 5000.00 |
+--------+---------+
3 rows in set (0.03 sec)

(root@localhost) [employees]>

第二步，自己join自己，条件为薪资=最大薪资且部门相同


(root@localhost) [employees]>select a.ename,a.deptno,a.sal from emp a join(select deptno,max(sal) as maxsal from emp group by deptno) b on a.sal=b.maxsal and a.deptno=b.deptno;
+-------+--------+---------+
| ename | deptno | sal     |
+-------+--------+---------+
| BLAKE |     30 | 2850.00 |
| SCOTT |     20 | 3000.00 |
| KING  |     10 | 5000.00 |
| FORD  |     20 | 3000.00 |
+-------+--------+---------+
4 rows in set (0.07 sec)

(root@localhost) [employees]>