
 1,查询所有
 1,查询所有
mysql> select * from grade;
 2,
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;
