MYSQL基础查询语言试题，请大家帮我做一下，谢谢了！ 200分！！

USE test

DROP TABLE if EXISTS grade

    create table `grade` (  

       `id` int PRIMARY KEY,        

       `name` varchar (300),           

       `score` double                  

    )   

    

    insert into `grade` (`id`, `name`, `score`) values('1','n1','59') 

    insert into `grade` (`id`, `name`, `score`) values('2','n2','66') 

    insert into `grade` (`id`, `name`, `score`) values('3','n3','78') 

    insert into `grade` (`id`, `name`, `score`) values('4','n1','48') 

    insert into `grade` (`id`, `name`, `score`) values('5','n3','85') 

    insert into `grade` (`id`, `name`, `score`) values('6','n5','51') 

    insert into `grade` (`id`, `name`, `score`) values('7','n4','98') 

    insert into `grade` (`id`, `name`, `score`) values('8','n5','53') 

    insert into `grade` (`id`, `name`, `score`) values('9','n2','67') 

    insert into `grade` (`id`, `name`, `score`) values('10','n4','88')

首先将上面内容保存在一个名字叫grade.sql 的文件里面，上面我私自修改了id的类型，实在看不下去了。然后登陆数据库使用mysql>source C:/grade.sql；

select name,max(score) from grade

select name,sum(score)  from grade group by name order by score desc limit 1,5

select name,sum(score) as tot from grade  group by name having tot<150

select name,avg(score) as scavg from grade group by name having scavg <80 and scavg >60

select name from grade group by name having sum(score) >150 and avg(score) <90这个应该是查询人吧，6才是查询人数吧

亲测有效

非常简单的： SELECT `class`, SUM(CASE WHEN score>=60 THEN 1 ELSE 0 END), SUM(CASE WHEN score>=60 THEN 0 ELSE 1 END) FROM tb1 GROUP BY 1 主要是考试GROUP BY分组统计，CASE WHEN THEN ELSE END分情形功能，这个是非常典型的应用，几乎所有的数据库都支持上面的语句。

---