MySQL 作业
一、create table grade(class int(20),chinese int(20),english int(20),math int(20),name varchar(20),age int(20),sid int(20) primary key);二、insert into grade(class,chinese,english,math,name,age,sid)values(1833,86,90,40,"zhangsan",21,1),(1832,55,86,66,"lisi",22,2),(1833,93,57,98,"zhaoliu",23,3),(1832,84,90,88,"wangwu",24,4),(1833,93,57,22,"lijiu",25,5),(1832,84,98,77,"niuqi",26,6),(1832,56,57,77,"liuli",27,7),(1833,48,58,88,"wangbo",28,8),(1832,78,57,88,"wangsan",29,9),(1833,87,60,65,"wanggan",30,10),(1832,80,76,88,"wangping",31,11),(1833,null,79,88,"wanghui",32,12); 1.select * from grade where class=1832;2.select * from grade where class=1833 and chinese between 80 and 90;3.select * from grade where sid limit 4,6;4.select name,sid from grade where class=1832 and english=98 and math=77;5.select * from grade where class=1832 order by chinese desc;6.select name from grade where class in(1833,1832) and chinese<80 and math<80;7.select name,class from grade where chinese is null;8.select name from grade where chinese<60 or chinese is null;9.select class,avg(math) from grade group by class;10.select class,sum(chinese) from grade group by class;11.update grade set chinese=60 where chinese<60 or chinese is null;12.select name,class from grade where chinese>70 and english>70 and math>70;13.select name,class from grade where english>70 and (chinese>60 or math>60);14.select class,count(*) from grade group by class;15.select class,count(*) from grade where math>80 group by class;16.17. alter table grade add(history int(20),biology int(20),geography int(20));file:///C:\Users\86176\AppData\Local\Temp\ksohtml\wps7489.tmp.jpg file:///C:\Users\86176\AppData\Local\Temp\ksohtml\wps748A.tmp.jpg
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