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文章目录

• 一、多表查询
• 1、链接查询
• • 1.1、SQL22 统计每个学校的答过题的用户的平均答题数
  • • 1.1.1、SQL语句(第一种写法)
    • 1.1.2、SQL语句(第二种写法)

一、多表查询

1、链接查询

1.1、SQL22 统计每个学校的答过题的用户的平均答题数

• 描述

运营想要了解每个学校答过题的用户平均答题数量情况，请你取出数据。

• 示例1

输入：
drop table if exists user_profile;
drop table if exists question_practice_detail;
CREATE TABLE user_profile (
device_id int NOT NULL,
gender varchar(14) NOT NULL,
age int ,
university varchar(32) NOT NULL,
gpa float,
active_days_within_30 int
);
CREATE TABLE question_practice_detail (
device_id int NOT NULL,
question_idint NOT NULL,
result varchar(32) NOT NULL
);

INSERT INTO user_profile VALUES(2138,‘male’,21,‘北京大学’,3.4,7);
INSERT INTO user_profile VALUES(3214,‘male’,null,‘复旦大学’,4.0,15);
INSERT INTO user_profile VALUES(6543,‘female’,20,‘北京大学’,3.2,12);
INSERT INTO user_profile VALUES(2315,‘female’,23,‘浙江大学’,3.6,5);
INSERT INTO user_profile VALUES(5432,‘male’,25,‘山东大学’,3.8,20);
INSERT INTO user_profile VALUES(2131,‘male’,28,‘山东大学’,3.3,15);
INSERT INTO user_profile VALUES(4321,‘male’,28,‘复旦大学’,3.6,9);
INSERT INTO question_practice_detail VALUES(2138,111,‘wrong’);
INSERT INTO question_practice_detail VALUES(3214,112,‘wrong’);
INSERT INTO question_practice_detail VALUES(3214,113,‘wrong’);
INSERT INTO question_practice_detail VALUES(6543,111,‘right’);
INSERT INTO question_practice_detail VALUES(2315,115,‘right’);
INSERT INTO question_practice_detail VALUES(2315,116,‘right’);
INSERT INTO question_practice_detail VALUES(2315,117,‘wrong’);
INSERT INTO question_practice_detail VALUES(5432,118,‘wrong’);
INSERT INTO question_practice_detail VALUES(5432,112,‘wrong’);
INSERT INTO question_practice_detail VALUES(2131,114,‘right’);
INSERT INTO question_practice_detail VALUES(5432,113,‘wrong’);

输出：
北京大学|1.0000
复旦大学|2.0000
山东大学|2.0000
浙江大学|3.0000

1.1.1、SQL语句(第一种写法)

select
u.university,
round(
(count(q.device_id) / count(distinct(q.device_id))),
4
) avg_answer_cnt
from
user_profile u,
question_practice_detail q
where
u.device_id = q.device_id
group by
university
order by
university asc

select  u.university,  round(    (count(q.device_id) / count(distinct(q.device_id))),    4  ) avg_answer_cntfrom  user_profile u,  question_practice_detail qwhere  u.device_id = q.device_idgroup by  universityorder by  university asc

1.1.2、SQL语句(第二种写法)

select
u.university,
round(
(count(q.device_id) / count(distinct(q.device_id))),
4
) avg_answer_cnt
from
user_profile u
join
question_practice_detail q
on
u.device_id = q.device_id
group by
university
order by
university asc

select  u.university,  round(    (count(q.device_id) / count(distinct(q.device_id))),    4  ) avg_answer_cntfrom  user_profile u  join  question_practice_detail qon  u.device_id = q.device_idgroup by  universityorder by  university asc

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