1，查询所有
mysql> select * from grade;
2，
mysql> select id,firstname,lastname from grade;
3，
mysql> select firstname,lastname from grade where id > 4;
4，
mysql> select * from grade where sex = 'f';
5，
mysql> select * from grade where id=2 or id =4 or id=6;
6，
mysql> select * from grade where math between 85 and 94;
7，
mysql> select * from grade where firstname like 'A%';
8，
mysql> select * from grade where firstname like 'A%e';
9，
mysql> select * from grade where firstname like '%l%';
10，
mysql> select * from grade where firstname like 'A____';
11，
mysql> select * from grade where english between 85 and 94;
12，
mysql> select * from grade where math >90 or chinese >90 ;
13，
mysql> select * from grade where id !=1 and id!=3 and id!= 5 and id!=7;
14，
mysql> select distinct sex from grade;
15，
mysql> select distinct lastname from grade;
16，
mysql> select count(id) from grade ;
17，
mysql> select count(id) from grade where english > 80 ;
18，
mysql> select sum(math) from grade;
19，
mysql> select sum(math) from grade where sex ='f';
20，
mysql> select avg(english) from grade;
21，
mysql> select avg(english) from grade where sex = 'm';
22，
mysql> select max(math) from grade;
23，
mysql> select max(math) from grade where sex='m';
24，
mysql> select * from grade order by math asc;
25，
mysql> select * from grade order by sex asc,chinese desc;
26，
mysql> select sex,GROUP_CONCAT(CONCAT(firstname, ' ', lastname) SEPARATOR ', ') AS students_list from grade group by sex;
27，
mysql> select count(*) from grade group by lastname;
28，
29，
mysql> select * from grade where id between 3 and 6;
30，
select id,sex,firstname,lastname,(english + math + chinese) / 3 AS average from grade where sex = 'm' AND (english + math + chinese) / 3 > 85;