【SQL每日一练】分组过滤练习题

十一*2024-04-22 15:09

文章目录

前言
MySQL语法
注意：

前言

题目：现在运营想查看每个学校用户的平均发贴和回帖情况，寻找低活跃度学校进行重点运营，请取出平均发贴数低于5的学校或平均回帖数小于20的学校。

复制代码

drop table if exists user_profile;
CREATE TABLE `user_profile` (
`id` int NOT NULL,
`device_id` int NOT NULL,
`gender` varchar(14) NOT NULL,
`age` int ,
`university` varchar(32) NOT NULL,
`gpa` float,
`active_days_within_30` int ,
`question_cnt` float,
`answer_cnt` float
);
INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4,7,2,12);
INSERT INTO user_profile VALUES(2,3214,'male',null,'复旦大学',4.0,15,5,25);
INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2,12,3,30);
INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6,5,1,2);
INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8,20,15,70);
INSERT INTO user_profile VALUES(6,2131,'male',28,'山东大学',3.3,15,7,13);
INSERT INTO user_profile VALUES(7,4321,'male',28,'复旦大学',3.6,9,6,52);

MySQL语法

代码如下：

复制代码

select university,AVG(question_cnt) as avg_question_cnt,AVG(answer_cnt) as avg_answer_cnt from user_profile 
group by university having avg_question_cnt< 5 or avg_answer_cnt <20

注意：

WHERE 关键字无法与合计函数一起使用；
SQL语句执行顺序

FROM - ON - JOIN - WHERE - GROUP BY - avg,sum.... - HAVING - SELECT - DISTINCT - ORDER BY - LIMIT

(8) SELECT (9) DISTINCT

(1) FROM

(3) JOIN

(2) ON

(4) WHERE

(5) GROUP BY

(6) avg,sum...

(7) HAVING

(10) ORDER BY

(11) LIMIT