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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。

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);

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MySQL语法

代码如下：

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 

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注意：

1. WHERE 关键字无法与合计函数一起使用；
2. 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