手撸Mysql原生语句--单表

select from where group by having order by limit

上面的所有操作是有执行的优先级的顺序的，我们将执行的过程可以总结为下面所示的七个步骤。

1.找到表:from

2.拿着where指定的约束条件，去文件/表中取出一条条记录

3.将取出的一条条记录进行分组group by，如果没有group by，则整体作为一组

4.执行select（去重）

5.将分组的结果进行having过滤

6.将结果按条件排序：order by

7.限制结果的显示条数

select from的使用

简单查询

SELECT id,emp_name,sex,age,hire_date,post,post_comment,salary,office,depart_id FROM employee;

SELECT * FROM employee;

SELECT emp_name,salary FROM employee;

避免重复DISTINCT

SELECT DISTINCT post FROM employee;

通过四则运算：

SELECT emp_name, salary*12 FROM employee;

SELECT emp_name, salary*12 AS Annual_salary FROM employee;

SELECT emp_name, salary*12 Annual_salary FROM employee;

1.查出所有员工的名字，薪资，格式为：<名字:egon> <薪资:3000>

select concat('<名字:',emp_name,'> ','<薪资:',salary,'>') from employee;

2.查询所有的岗位（去掉重复的岗位）

select distinct depart_id from employee;

3.查询所有员工的名字以及他们的年薪，年薪的字段名为annual_year

select emp_name, salary*12 annual_year from employee;

where约束

1.查看岗位是teacher的员工的姓名、年龄

select emp_name,age from emp_name where post = 'teacher';

2.查看岗位是teacher且年龄大于30岁的员工的姓名、年龄

select emp_name,age from employee where post ='teacher' and age >30;

3.查看岗位是teacher且薪资在9000-10000范围内的员工姓名、年龄、薪资

select emp_name,age,salary from employee where post ='teacher' and salary between 9000 and 10000;

select emp_name,age,salary from employee where post ='teacher' and salary >=9000 and salary <=10000;

4.查看岗位描述不为null的员工信息

select * from employee where post_comment is not null;

5.查看岗位是teacher且薪资是10000或9000或30000的员工姓名、年龄、薪资。

select emp_name,age,salary from employee where post = 'teacher' and salary in (10000,9000,30000);

6.查看岗位是teacher且薪资不是10000或9000或30000的员工姓名、年龄、薪资。

select emp_name,age,salary from employee where post='teacher' and salary not in (10000,9000,30000);

7.查看岗位是teacher且名字是jin开头的员工姓名、年薪

select emp_name,salary*12 annual_year from employee where post = 'teacher' and emp_name like 'jin%';

select emp_name,salary*12 annual_year from employee where post = 'teacher' and emp_name regexp '^jin';

group by--聚合(sum,avg,min,max,count)的使用

1.查询岗位名及岗位包含的所有员工名字

select post, group_concat(emp_name) from employee group by post;

select post, group_concat(emp_name) as emp_members from employee group by post;

2.查询岗位名以及岗位内包含的员工个数

select post,count(id) from employee group by post;

3.查询公司内男员工和女员工的个数

select sex,count(id) from employee group by sex;

4.查询岗位名以及各岗位的平均薪资

select post,avg(salary) from employee group by post;

5.查询岗位名以及各个岗位的最高薪资

select post,max(salary) from employee group by post;

6.查询敢为名以及各个岗位的最低信息

select post,min(salary) from employee group by post;

7.查询男员工与男员工的平均薪资，女员工和女员工的平均薪资

select sex,avg(salary) from employee group by sex;

8.查询岗位名以及岗位包含员工的所有薪资的总和

select post,sum(salary) from employee group by post;

having过滤的使用

！！！执行优先级从高到低：where > group by > having

1. Where 发生在分组group by之前，因而Where中可以有任意字段，但是绝对不能使用聚合函数。
2. Having发生在分组group by之后，因而Having中可以使用分组的字段，无法直接取到其他字段,可以使用聚合函数

1.查询各岗位内包含的员工个数小于2的岗位名、岗位内包含员工名字、个数

select post,group_concat(emp_name),count(id) from employee group by post having count(id)<2;

2.查询各岗位平均薪资大于10000的岗位名、平均工资

select post,avg(salary) from employee group by post having avg(salary) >10000;

3.查询各岗位平均薪资大于10000且小于20000的岗位名、平均薪资

select post,avg(salary) from employee group by post having avg(salary) between 10000 and 20000;

select post,avg(salary) from employee group by post having avg(salary) > 10000 and avg(salary) <20000;

order by查询排序的使用

1. 查询所有员工信息，先按照age升序排序，如果age相同则按照hire_date降序排序
   
   select * from employee order by age asc,hire_date desc;

2. 查询各岗位平均薪资大于10000的岗位名、平均工资,结果按平均薪资升序排列
   
   select post,avg(salary) from employee group by post having avg(salary)>10000 order by avg(salary) asc;

3. 查询各岗位平均薪资大于10000的岗位名、平均工资,结果按平均薪资降序排列
   
   select post,avg(salary) from employee group by post having avg(salary)>10000 order by avg(salary) desc;

limit限制查询的记录数

select * from employee limit 0,5;

select * from employee limit 5,5;

select * from employee limit 10,5;

备注：limit后面的两个数字分别表示的含义为：第一个数字表示的含义是从第几条数据开始，第二个数字表示的是需要取出几条的数据。

使用正则表达式查询

select * from employee where emp_name REGEXP '^ale';

SELECT * FROM employee WHERE emp_name REGEXP 'on$';

SELECT * FROM employee WHERE emp_name REGEXP 'm{2}';

小结：对字符串匹配的方式

WHERE emp_name = 'egon';

WHERE emp_name LIKE 'yua%';

WHERE emp_name REGEXP 'on$';

1.查看所有员工中名字是jin开头，n或者g结果的员工信息。

select * from employee where emp_name regexp '^jin.*[ng]$';