【MySQL】基本查询（插入查询结果、聚合函数、分组查询）

一、插入查询结果

语法：

INSERT INTO table_name [(column [, column ...])] SELECT ...

案例：删除表中重复数据

--创建初始重复表
mysql> CREATE TABLE duplicate_table (id int, name varchar(20));
Query OK, 0 rows affected (0.03 sec)
--插入重复数据
mysql> INSERT INTO duplicate_table VALUES
    -> (100, 'aaa'),
    -> (100, 'aaa'),
    -> (200, 'bbb'),
    -> (200, 'bbb'),
    -> (200, 'bbb'),
    -> (300, 'ccc');
Query OK, 6 rows affected (0.00 sec)
Records: 6  Duplicates: 0  Warnings: 0
--查询表中数据
mysql> select * from duplicate_table;
+------+------+
| id   | name |
+------+------+
|  100 | aaa  |
|  100 | aaa  |
|  200 | bbb  |
|  200 | bbb  |
|  200 | bbb  |
|  300 | ccc  |
+------+------+
6 rows in set (0.00 sec)
--新建一个相同表结构的空表
mysql> create table no_duplicate_table like duplicate_table;
Query OK, 0 rows affected (0.02 sec)
--把去重后的结果插入空表中
mysql> insert into no_duplicate_table select distinct *from duplicate_table;
Query OK, 3 rows affected (0.01 sec)
Records: 3  Duplicates: 0  Warnings: 0
//查询表内数据
mysql> select * from no_duplicate_table;
+------+------+
| id   | name |
+------+------+
|  100 | aaa  |
|  200 | bbb  |
|  300 | ccc  |
+------+------+
3 rows in set (0.00 sec)
//修改两个表名，把去重后的表该为该名字
mysql> rename table duplicate_table to old_duplicate_table,no_duplicate_table to duplicate_table;
Query OK, 0 rows affected (0.02 sec)

mysql> select * from duplicate_table;
+------+------+
| id   | name |
+------+------+
|  100 | aaa  |
|  200 | bbb  |
|  300 | ccc  |
+------+------+
3 rows in set (0.00 sec)

这里通过rename修改表名是为了等表的操作结束后，统一放入，更新，生效，节省时间

二、聚合函数

MySQL中的聚合函数常用于对数据进行计算和统计，以下是几种常见的聚合函数

案例：

• 统计班级共有多少同学

mysql> select * from exam_result;
+----+-----------+---------+------+---------+
| id | name      | chinese | math | english |
+----+-----------+---------+------+---------+
|  1 | 唐三藏    |     134 |   98 |      56 |
|  3 | 猪悟能    |     176 |   98 |      90 |
|  4 | 曹孟德    |     140 |   90 |      67 |
|  5 | 刘玄德    |     110 |  115 |      45 |
|  6 | 孙权      |     140 |   73 |      78 |
|  7 | 宋公明    |     150 |   95 |      30 |
+----+-----------+---------+------+---------+
6 rows in set (0.00 sec)

mysql> select count(*) from exam_result;
+----------+
| count(*) |
+----------+
|        6 |
+----------+
1 row in set (0.00 sec)

mysql> select count(1) from exam_result;
+----------+
| count(1) |
+----------+
|        6 |
+----------+
1 row in set (0.00 sec)

-- 统计班级的数学成绩有多少个（去重）
mysql> select math  from exam_result;
+------+
| math |
+------+
|   98 |
|   98 |
|   90 |
|  115 |
|   73 |
|   95 |
+------+
6 rows in set (0.01 sec)

mysql> select count(distinct math) from exam_result;
+----------------------+
| count(distinct math) |
+----------------------+
|                    5 |
+----------------------+
1 row in set (0.00 sec)

-- 统计数学成绩总分
mysql> select sum(math) from exam_result;
+-----------+
| sum(math) |
+-----------+
|       569 |
+-----------+
1 row in set (0.00 sec)

--统计数学成绩的平均分
mysql> select avg(math) from exam_result;
+-------------------+
| avg(math)         |
+-------------------+
| 94.83333333333333 |
+-------------------+
1 row in set (0.00 sec)

--统计英语成绩不及格的人数
mysql> select count(*) from exam_result where english<60;
+----------+
| count(*) |
+----------+
|        3 |
+----------+
1 row in set (0.00 sec)

--返回英语最高分
mysql> select max(english) from exam_result;
+--------------+
| max(english) |
+--------------+
|           90 |
+--------------+
1 row in set (0.00 sec)

--返回 > 70 分以上的数学最低分
mysql> select min(math) from exam_result where math>70;
+-----------+
| min(math) |
+-----------+
|        73 |
+-----------+
1 row in set (0.00 sec)

三、分组查询（group by & having）

分组的目的是为了方便进行聚合统计

在select中使用group by 子句可以对指定列进行分组查询

select column1, column2, .. from table group by column;

案例：
EMP员工表
DEPT部门表
SALGRADE工资等级表

1. 显示每个部门的平均工资和最高工资

group by ‘列名’：分组是以同一列不同行数据来进行分组的；分组过后，每组内的【分组列名如deptno】，一定是一样的，可以被聚合压缩

mysql> select deptno,avg(sal) 平均工资, max(sal) '最高工资' from emp group by deptno;
+--------+--------------+--------------+
| deptno | 平均工资     | 最高工资     |
+--------+--------------+--------------+
|     10 |  2916.666667 |      5000.00 |
|     20 |  2175.000000 |      3000.00 |
|     30 |  1566.666667 |      2850.00 |
+--------+--------------+--------------+
3 rows in set (0.00 sec)

2. 显示每个部门的每种岗位的平均工资和最低工资

mysql> select deptno, job,avg(sal) 平均工资, min(sal)最低工资 from emp group by deptno, job;
+--------+-----------+--------------+--------------+
| deptno | job       | 平均工资     | 最低工资     |
+--------+-----------+--------------+--------------+
|     10 | CLERK     |  1300.000000 |      1300.00 |
|     10 | MANAGER   |  2450.000000 |      2450.00 |
|     10 | PRESIDENT |  5000.000000 |      5000.00 |
|     20 | ANALYST   |  3000.000000 |      3000.00 |
|     20 | CLERK     |   950.000000 |       800.00 |
|     20 | MANAGER   |  2975.000000 |      2975.00 |
|     30 | CLERK     |   950.000000 |       950.00 |
|     30 | MANAGER   |  2850.000000 |      2850.00 |
|     30 | SALESMAN  |  1400.000000 |      1250.00 |
+--------+-----------+--------------+--------------+
9 rows in set (0.00 sec)

注意事项：在group by之后出现的字段是可以在select 之后出现的，还有聚合函数，正常分组出现的字段在聚合条件中可以输出，其他会报错

select ename,deptno,job,avg(sal)平均,min(sal) 最低 from emp group by deptno,job;

上面的代码因为分组条件中没有用到ename 所以报错

3. 显示平均工资低于2000的部门和它的平均工资

mysql> select deptno,avg(sal) deptavg from emp group by deptno having deptavg<2000;
+--------+-------------+
| deptno | deptavg     |
+--------+-------------+
|     30 | 1566.666667 |
+--------+-------------+
1 row in set (0.00 sec)

4. 除SMITH外，显示平均工资低于2000的每个部门的每种岗位的和它的平均工资

mysql> select deptno,job,avg(sal) deptavg from emp where ename!='SMITH' group by deptno,job having deptavg<2000;
+--------+----------+-------------+
| deptno | job      | deptavg     |
+--------+----------+-------------+
|     10 | CLERK    | 1300.000000 |
|     20 | CLERK    | 1100.000000 |
|     30 | CLERK    |  950.000000 |
|     30 | SALESMAN | 1400.000000 |
+--------+----------+-------------+
4 rows in set (0.00 sec)

四、SQL查询的执行顺序

SQL查询中各个关键字的执行先后顺序 ：from > on> join > where > group by > with > having > select>distinct > order by > limit

五、OJ练习

1.批量插入数据
答案：

insert into actor values(1,'PENELOPE','GUINESS','2006-02-15 12:34:33'),(2,'NICK','WAHLBERG','2006-02-15 12:34:33');

2.找出所有员工薪水情况
答案：

select distinct salary from salaries order by salary desc;

3.查找最晚入职员工的所有信息

答案：

select * from employees order by hire_date desc limit 1;

4.查找入职员工时间排名倒数第三的员工所有信息

答案：

select * from employees where hire_date=(select distinct hire_date from employees order by hire_date desc limit 2,1);

5.查找薪水记录超过15条的员工号emp_no以及其对应的记录次数t
分组+聚合函数

答案：

select emp_no,count(*) t from salaries group by emp_no having t>15;

6.获取所有部门薪水
答案：

SELECT dm.dept_no, dm.emp_no, s.salary
FROM dept_manager dm
JOIN salaries s ON dm.emp_no = s.emp_no
WHERE dm.to_date = '9999-01-01' AND s.to_date = '9999-01-01'
ORDER BY dm.dept_no ASC;

--或者
SELECT
    dm.dept_no,
    dm.emp_no,
    (
        SELECT s.salary
        FROM salaries s
        WHERE s.emp_no = dm.emp_no
        AND s.to_date = '9999-01-01'
        LIMIT 1
    ) AS salary
FROM
    dept_manager dm
WHERE
    dm.to_date = '9999-01-01'
ORDER BY
    dm.dept_no ASC;

7.从titles表获取按照title进行分组

答案：

select title,count(title) t from titles group by title having t>=2; 

8.查找重复数据

答案

select email from Person group by email having count(email)>1;

9.查找大国

select name,population,area from World where area>=3000000 or population>=25000000;

10.给定一个Employee表，要找出其中第N高的薪资（Salary）

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  SET N = N - 1;
  RETURN (
     
      select distinct(Salary) as getNthHighestSalary
      from Employee
      GROUP BY Salary 
      ORDER BY Salary DESC 
      limit 1 offset N
  );
END