数据库SQL练习 - 废江's博客废江's博客

文章目录[隐藏]

12-8
12-10
1-4

12-8

题目：统计每个学校的答过题的用户的平均答题数

咋一看无从下手，其实很简单。每个学校，则说明按学校分组，平均答题数，则是一个学校的所有学生的答题总数/学生总数。注意前缀，我加了一个学校的，那么按学校分组也是刚刚好，注意一点就是学生总数需要去重，题目总数不必去重

select university,
count(question_id) / count(distinct(q.device_id)) as asavg_answer_cnt
from user_profile u join question_practice_detail q on u.device_id = q.device_id
group by university

select university,

count(question_id) / count(distinct(q.device_id)) as asavg_answer_cnt

from user_profile u join question_practice_detail q on u.device_id = q.device_id

group by university

题目：统计每个学校各难度的用户平均刷题数

从题目分析，这是多分组条件，即根据学校和题目难度两个条件分组，然后求用户平均刷题数,也就是用户的回答题目总数/用户总数。用户总数需要去重

select university,difficult_level,
count(answer_cnt) / count(DISTINCT u.device_id) as avg_answer_cnt
from user_profile u,question_practice_detail qpd,question_detail qd
where u.device_id = qpd.device_id and qpd.question_id = qd.question_id
group by university,difficult_level

select university,difficult_level,

count(answer_cnt) / count(DISTINCT u.device_id) as avg_answer_cnt

from user_profile u,question_practice_detail qpd,question_detail qd

where u.device_id = qpd.device_id and qpd.question_id = qd.question_id

group by university,difficult_level

这里，三张表连接join我不太会了，参考一下下面的表连接join查询方式

SELECT
    university,
    difficult_level,
    count(q.question_id )/count(
    distinct(q.device_id)) avg_answer_cnt 
FROM
    user_profile u
    JOIN question_practice_detail q ON u.device_id = q.device_id
    JOIN question_detail qd ON q.question_id = qd.question_id 
GROUP BY
    university,
    difficult_level;

SELECT

university,

difficult_level,

count(q.question_id )/count(

distinct(q.device_id)) avg_answer_cnt

FROM

user_profile u

JOIN question_practice_detail q ON u.device_id = q.device_id

JOIN question_detail qd ON q.question_id = qd.question_id

GROUP BY

university,

difficult_level;

题目：统计每个用户的平均刷题数

现在做这题目也就是得心应手了，山东大学的用户在不同难度下的平均答题题目数嘛，分组嘛，根据难度分，where过滤山东大学，平均答题数则是总答题数/总人数

select university,difficult_level,
count(qpd.question_id) / count(distinct qpd.device_id) avg_answer_cnt
from user_profile u 
join question_practice_detail qpd on u.device_id = qpd.device_id
join question_detail qd on qpd.question_id = qd.question_id
where university = "山东大学"
group by difficult_level

select university,difficult_level,

count(qpd.question_id) / count(distinct qpd.device_id) avg_answer_cnt

from user_profile u

join question_practice_detail qpd on u.device_id = qpd.device_id

join question_detail qd on qpd.question_id = qd.question_id

where university = "山东大学"

group by difficult_level

12-10

题目：计算25岁以上和以下的用户数量

这里，考了一个新知识点，之前我是没见过的：

条件函数if。if（x=n,a,b）表示如果x=n,则返回a，否则就是b了。

select age_cut,count(device_id)as number 
from(Select if(age>=25,'25岁及以上','25岁以下' )as age_cut,device_id 
From user_profile)u2
Group by age_cut

select age_cut,count(device_id)as number

from(Select if(age>=25,'25岁及以上','25岁以下' )as age_cut,device_id

From user_profile)u2

Group by age_cut

但仔细分析题目，就是有一个明显特点，就是我可以改成两句，查询！没错，所有也可以使用联合查询来实现

select '25岁以下' as age_cut,count(device_id) as number
from user_profile
where age<25 or age is null
union all
select '25岁及以上' as age_cut,count(device_id) as number
from user_profile
where age>=25;

select '25岁以下' as age_cut,count(device_id) as number

from user_profile

where age<25 or age is null

union all

select '25岁及以上' as age_cut,count(device_id) as number

from user_profile

where age>=25;

题目：查看不同年龄段的用户明细

使用case来查询即可，掌握case 的用法

SELECT device_id,gender,
       (CASE
            WHEN age >= 25 THEN '25岁及以上'
            WHEN age BETWEEN 20 AND 24 THEN '20-24岁'
            WHEN age < 20 THEN '20岁以下'
            ELSE '其他'
        END) AS age_cut
FROM user_profile;

SELECT device_id,gender,

(CASE

WHEN age >= 25 THEN '25岁及以上'

WHEN age BETWEEN 20 AND 24 THEN '20-24岁'

WHEN age < 20 THEN '20岁以下'

ELSE '其他'

END) AS age_cut

FROM user_profile;

题目：计算用户8月每天的练题数量

mysql中的日期是可以分隔的，可能你看到了2021-08-15，陷入了深思，如何分组根据2021-08.其实，mysql早就想好了，可以只查询8月份的，然后分组

Select day(date) as day, count(question_id) as question_cnt
From question_practice_detail
Where year(date)=2021 and month(date)=08
Group by day

Select day(date) as day, count(question_id) as question_cnt

From question_practice_detail

Where year(date)=2021 and month(date)=08

Group by day

题目：计算用户的平均次日留存率

太难，放个答案。思路是计算前一天的用户刷题数量，当天的用户刷题数量，然后相除

select avg(if(b.device_id is not null,1,0)) as avg_ret
from 
(select distinct device_id,date
from question_practice_detail
)a
left join 
(
select distinct device_id,date_sub(date,interval 1 day) as date 
from question_practice_detail
)b
on a.device_id = b.device_id and a.date = b.date

select avg(if(b.device_id is not null,1,0)) as avg_ret

from

(select distinct device_id,date

from question_practice_detail

left join

(

select distinct device_id,date_sub(date,interval 1 day) as date

from question_practice_detail

on a.device_id = b.device_id and a.date = b.date

题目：统计每种性别的人数

1、LOCATE(substr , str )：返回子串 substr 在字符串 str 中第一次出现的位置，如果字符substr在字符串str中不存在，则返回0；

2、POSITION(substr IN str )：返回子串 substr 在字符串 str 中第一次出现的位置，如果字符substr在字符串str中不存在，与LOCATE函数作用相同；

3、LEFT(str, length)：从左边开始截取str，length是截取的长度；

4、RIGHT(str, length)：从右边开始截取str，length是截取的长度；

5、SUBSTRING_INDEX(str ,substr ,n)：返回字符substr在str中第n次出现位置之前的字符串;

6、SUBSTRING(str ,n ,m)：返回字符串str从第n个字符截取到第m个字符；

7、REPLACE(str, n, m)：将字符串str中的n字符替换成m字符；

8、LENGTH(str)：计算字符串str的长度。

select substring_index(profile,',',-1) as gender,count(device_id)
from user_submit
group by gender

select substring_index(profile,',',-1) as gender,count(device_id)

from user_submit

group by gender

1-4

题目：截取年龄

考察字符串的练习，对函数substring_index和函数substring的掌握

select substring_index(SUBSTRING_INDEX(profile,",",3),",",-1) as age,count(device_id)
FROM user_submit
group by age

select substring_index(SUBSTRING_INDEX(profile,",",3),",",-1) as age,count(device_id)

FROM user_submit

group by age

题目：找出每个学校GPA最低的同学

如果只是找出gpq最低的，那么很简单，直接分组，select min（gpa）就行了，但是现在还要关联学校，我之前写的一个答案是错的：

错误的：

select device_id,university,gpa
FROM user_profile
WHERE gpa IN(
select min(gpa) 
from user_profile
group by university
    )
GROUP BY university #保证学校名字不重复 
order BY university #保证与题目要求输出一致

select device_id,university,gpa

FROM user_profile

WHERE gpa IN(

select min(gpa)

from user_profile

group by university

)

GROUP BY university #保证学校名字不重复

order BY university #保证与题目要求输出一致

这个错在哪里？我就不多说了，你肯定需要对某一行数据进行关联，看下面的正确答案

正确的：

SELECT
	device_id,
	university,
	gpa 
FROM
	user_profile u 
WHERE
	gpa = ( SELECT MIN( gpa ) FROM user_profile WHERE university = u.university ) 
ORDER BY
	university

SELECT

device_id,

university,

gpa

FROM

user_profile u

WHERE

gpa = ( SELECT MIN( gpa ) FROM user_profile WHERE university = u.university )

ORDER BY

university

题目：统计复旦用户8月练题情况

表连接，然后筛选即可

SELECT
    u.device_id,
    university,
    SUM(IF(result IS NOT NULL, 1, 0)) AS questino_cnt,
    SUM(IF(result = "right", 1, 0)) AS right_question_cnt
FROM
    user_profile u
    LEFT JOIN question_practice_detail q ON u.device_id = q.device_id
    AND MONTH(q.`date`) = "08"
WHERE 
    university = "复旦大学"
GROUP BY
    u.device_id;

SELECT

u.device_id,

university,

SUM(IF(result IS NOT NULL, 1, 0)) AS questino_cnt,

SUM(IF(result = "right", 1, 0)) AS right_question_cnt

FROM

user_profile u

LEFT JOIN question_practice_detail q ON u.device_id = q.device_id

AND MONTH(q.`date`) = "08"

WHERE

university = "复旦大学"

GROUP BY

u.device_id;

题目：浙大不同难度题目的正确率

很难，但是得去想，我先把正确率求出来吧？不看浙大学校，但我遇到了困难

如何对一个字段中进行统计啊？这里我只会拿到result中的数量，我不会拿到当他等于right的数量，然后相除得到正确率。

SELECT difficult_level,COUNT(result)
FROM question_detail as qd,question_practice_detail as qpd
WHERE qd.question_id = qpd.question_id
GROUP BY difficult_level

SELECT difficult_level,COUNT(result)

FROM question_detail as qd,question_practice_detail as qpd

WHERE qd.question_id = qpd.question_id

GROUP BY difficult_level

后来，我会了，原来是使用if关键字….我会了，可她已经不在了…啊偏题了，拿到正确率，接下来就是绑定浙江大学

SELECT difficult_level,sum(if(result = 'right', 1, 0)) / count(*) as correct_rate
FROM question_detail as qd,question_practice_detail as qpd
WHERE qd.question_id = qpd.question_id
GROUP BY difficult_level

SELECT difficult_level,sum(if(result = 'right', 1, 0)) / count(*) as correct_rate

FROM question_detail as qd,question_practice_detail as qpd

WHERE qd.question_id = qpd.question_id

GROUP BY difficult_level

最后很简单了，只需要关联浙江大学，但是tmd结果还是错了，

SELECT difficult_level,sum(if(result = 'right', 1, 0)) / count(*) as correct_rate
FROM 
question_detail as qd,
question_practice_detail as qpd,
user_profile as up
WHERE 
qd.question_id = qpd.question_id
and
up.device_id = qpd.device_id
and
university = "浙江大学"
GROUP BY difficult_level

SELECT difficult_level,sum(if(result = 'right', 1, 0)) / count(*) as correct_rate

FROM

question_detail as qd,

question_practice_detail as qpd,

user_profile as up

WHERE

qd.question_id = qpd.question_id

and

up.device_id = qpd.device_id

and

university = "浙江大学"

GROUP BY difficult_level

别慌，题目么看清，加个升序就行了。

总结，对于这种难sql，可以先写一部分拆分，但是有些知识不会，怎么拆都写不出来，比如先前的if，就得记住，得会。拆完之后，绑定浙江大学就十分的轻松了

题目：21年8月份练题总数

select count(DISTINCT device_id),count(question_id) from question_practice_detail
where date like "2021-08%"

1 2	select count(DISTINCT device_id),count(question_id) from question_practice_detail where date like "2021-08%"

到此，基本练习的题目都写完了，加油！

一	二	三	四	五	六	日
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30

12-8

题目：统计每个学校的答过题的用户的平均答题数

题目：统计每个学校各难度的用户平均刷题数

题目：统计每个用户的平均刷题数

12-10

题目：计算25岁以上和以下的用户数量

题目：查看不同年龄段的用户明细

题目：计算用户8月每天的练题数量

题目：计算用户的平均次日留存率

题目：统计每种性别的人数

1-4

题目：截取年龄

题目：找出每个学校GPA最低的同学

错误的：

正确的：

题目：统计复旦用户8月练题情况

题目：浙大不同难度题目的正确率

总结，对于这种难sql，可以先写一部分拆分，但是有些知识不会，怎么拆都写不出来，比如先前的if，就得记住，得会。拆完之后，绑定浙江大学就十分的轻松了

题目：21年8月份练题总数

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