SELECT DISTINCT 字段1,字段2... FROM 表名WHERE 条件GROUP BY fieldHAVING 筛选ORDER BY fieldLIMIT 限制条数

from
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1.找到表 :from2.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.这行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 by7.限制结果的显示条数

#简单查询
select 字段名字,字段名字 ,字段名字 from 表名   ----名字查询select * from 表名   -----查询所有#避免重复DISTINCT
select distinct 字段名字 from 表名;#定义显示格式

concat() 函数用于连接字符串
select concat('姓名:',字段名) from 表名;concat_ws() 第一个参数为分隔线
select concat_ws(':',字段名) from 表名;#通过四则运算查询
select 字段名, salary*12(月薪*12) as annual_salary(年薪) from 表名#运算加重新定义名字

#单条件查询select 字段名 from 表名 where 条件;
例:select emp_name from employee where post='sale';#多条件查询

select 字段名 from  employee where 条件 and 条件例:SELECT emp_name,salary FROM employee WHERE post='teacher' AND salary>10000;#关键字between and
例: SELECT emp_name,salary FROM employee WHERE salary BETWEEN 10000 AND 20000;SELECT emp_name,salary FROM employee WHERE salary NOT BETWEEN 10000 AND 20000;#关键字 IS NULL(判断某个字段是否为null不能用等号,需要用IS)
例:SELECT emp_name,post_comment FROM employee WHERE post_comment IS NULL;SELECT emp_name,post_comment FROM employee WHERE post_comment IS NOT NULL;
注意是空字符串,不是null#关键字IN集合查询
例:SELECT emp_name,salary FROM employee WHERE salary=3000 OR salary=3500 OR salary=4000 OR salary=9000 ;SELECT emp_name,salary FROM employee WHERE salary IN (3000,3500,4000,9000) ;SELECT emp_name,salary FROM employee WHERE salary NOT IN (3000,3500,4000,9000) ;
#关键字like模糊查询
例:  通配符’%’SELECT * FROM employee WHERE emp_name LIKE 'eg%';通配符’_’SELECT * FROM employee WHERE emp_name LIKE 'al__';

#单独使用group by 关键字分组
select post from employee group by post;#group by关键字和group_concat()函数一起使用

select post,group_concat(emp_name) from employee group by post;#按照岗位分组,并查询组内成员名字#group by 与聚合函数一起使用
select post,count(id) from employee group by post;#按照岗位分组,并查看每个有多少人'''
强调:
如果我们用unique的字段作为分组的依据，则每一条记录自成一组，这种分组没有意义
多条记录之间的某个字段值相同，该字段通常用来作为分组的依据
'''

#强调：聚合函数聚合的是组的内容，若是没有分组，则默认一组

示例：SELECT COUNT(*) FROM employee;SELECT COUNT(*) FROM employee WHERE depart_id=1;SELECT MAX(salary) FROM employee;SELECT MIN(salary) FROM employee;SELECT AVG(salary) FROM employee;SELECT SUM(salary) FROM employee;SELECT SUM(salary) FROM employee WHERE depart_id=3;

HAVING与WHERE不一样的地方在于!!!!!!#！！！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

mysql> select @@sql_mode;
+--------------------+
| @@sql_mode         |
+--------------------+
| ONLY_FULL_GROUP_BY |
+--------------------+
row in set (0.00 sec)mysql> select * from emp where salary > 100000;
+----+------+------+-----+------------+---------+--------------+------------+--------+-----------+
| id | emp_name | sex  | age | hire_date  | post    | post_comment | salary     | office | depart_id |
+----+------+------+-----+------------+---------+--------------+------------+--------+-----------+
|  2 | alex | male |  78 | 2015-03-02 | teacher | NULL         | 1000000.31 |    401 |         1 |
+----+------+------+-----+------------+---------+--------------+------------+--------+-----------+
row in set (0.00 sec)mysql> select post,group_concat(emp_name) from emp group by post having salary > 10000;#错误，分组后无法直接取到salary字段
ERROR 1054 (42S22): Unknown column 'salary' in 'having clause'
mysql> select post,group_concat(emp_name) from emp group by post having avg(salary) > 10000;
+-----------+-------------------------------------------------------+
| post | group_concat(emp_name) |
+-----------+-------------------------------------------------------+
| operation | 程咬铁,程咬铜,程咬银,程咬金,张野 |
| teacher | 成龙,jinxin,jingliyang,liwenzhou,yuanhao,wupeiqi,alex |
+-----------+-------------------------------------------------------+
rows in set (0.00 sec)验证

验证

按单列排序SELECT * FROM employee ORDER BY salary;SELECT * FROM employee ORDER BY salary ASC;  #升序SELECT * FROM employee ORDER BY salary DESC;#降序

按多列排序:先按照age排序，如果年纪相同，则按照薪资排序SELECT * from employeeORDER BY age,salary DESC;

示例：SELECT * FROM employee ORDER BY salary DESC LIMIT 3;                    #默认初始位置为0
    SELECT * FROM employee ORDER BY salary DESCLIMIT 0,5; #从第0开始，即先查询出第一条，然后包含这一条在内往后查5条
SELECT * FROM employee ORDER BY salary DESCLIMIT 5,5; #从第5开始，即先查询出第6条，然后包含这一条在内往后查5条

SELECT * FROM employee WHERE emp_name REGEXP '^ale';   #以ale开头SELECT * FROM employee WHERE emp_name REGEXP 'on$';   #以on结尾SELECT * FROM employee WHERE emp_name REGEXP 'm{2}';   #连续的两个m