【SQL进阶】【分步写、联合各自排序、TIMESTAMPDIFF时间比较】Day04：多表查询

〇、内容

时间比较2-2

联合结果各自排序

查询列和GROUP BY

一、嵌套子查询

1、月均完成试卷数不小于3的用户爱作答的类别

 

自己的答案【错误】：

SELECT tag,
    COUNT(A.start_time) AS tag_cnt
FROM (
    -- 查询 “当月均完成试卷数”不小于3的用户们
    SELECT *
    FROM exam_record
    GROUP BY uid
    HAVING COUNT(*)>=3
) A
RIGHT JOIN examination_info B
ON A.exam_id=B.exam_id    
GROUP BY tag
ORDER BY tag_cnt DESC

答案：【group by的字段一定要出现在查询列中，*不算】

SELECT tag,
    COUNT(B.tag) AS tag_cnt
FROM exam_record A
RIGHT JOIN examination_info B
ON A.exam_id=B.exam_id
WHERE uid IN (
    SELECT uid
    FROM exam_record
    GROUP BY uid
    -- 统计当前用户完成试卷总数
    -- 统计该用户有完成试卷的月份数
    HAVING COUNT(submit_time) / COUNT(DISTINCT DATE_FORMAT(submit_time, "%Y%m")) >= 3
)
GROUP BY tag
ORDER BY tag_cnt DESC

2、试卷发布当天作答人数和平均分【☆】

 

 自己的答案【错误】

SELECT
    exam_id,
    SUM(IF(DATE_FORMAT((submit_time,"%Y%m")==DATE_FORMAT(release_time,"%Y%m") and level>=5,1,0))) AS uv,
    ROUND(AVG(score),1) AS avg_score
FROM user_info A
JOIN examination_info
JOIN exam_record
ON 
    A.uid=C.uid
    AND
    B.exam_id=C.exam_id
WHERE 
    tag="SQL"    
GROUP BY exam_id
ORDER BY uv DESC,avg_score ASC

正确答案：【判断相等用一个等号】

SELECT
    C.exam_id,
    COUNT(DISTINCT C.uid) AS uv,
    ROUND(AVG(score),1) AS avg_score
FROM user_info A
JOIN examination_info B
JOIN exam_record C
ON 
    A.uid=C.uid
    AND
    B.exam_id=C.exam_id
WHERE 
    tag="SQL"
    AND
    level>5
    AND
    DATE(submit_time)=DATE(release_time)
GROUP BY C.exam_id
ORDER BY uv DESC,avg_score ASC

3、作答试卷得分大于过80的人的用户等级分布

 

SELECT
    level,
    COUNT(*) AS level_cnt
FROM user_info A
JOIN examination_info B
JOIN exam_record C
-- NATURAL/FULL/CROSS
ON A.uid=C.uid
AND B.exam_id=C.exam_id
WHERE 
    tag="SQL"
    AND
    score>80
GROUP BY level
ORDER BY level_cnt DESC

二、合并查询

1、每个题目和每份试卷被作答的人数和次数

 

答案：【UNION和ORDER BY混用会被覆盖】

-- order by可以存在 union的字句里面，但是功能不会生效
SELECT * FROM
(SELECT
    exam_id AS tid,
    COUNT(DISTINCT uid) AS uv,
    COUNT(exam_id) AS pv
FROM exam_record
GROUP BY tid
ORDER BY uv DESC,pv DESC) AS A
UNION
SELECT * FROM 
(SELECT
    question_id AS tid,
    COUNT(DISTINCT uid) AS uv,
    COUNT(question_id) AS pv
FROM practice_record
GROUP BY tid
ORDER BY uv DESC,pv DESC) AS B

2、分别满足两个活动的人

思路：TIMESTAMPDIFF(SECOND,start_time,submit_time)<=duration*30，时间比较用SECOND和TIMESTAMPDIFF

至少有一次不需要聚合函数，在where中即可实现，先查找出符合要求的，再进行分组

全部需要用聚合函数，先分组再having

-- 输出2021年里，所有每次试卷得分都能到85分的人以及至少有一次
-- 用了一半时间就完成高难度试卷且分数大于80的人的id和活动号，按用户ID排序输出。
-- 全部成绩大于85可以用最小成绩>85表示
(SELECT
    uid,
    "activity1" AS activity
FROM exam_record
GROUP BY uid
HAVING MIN(score)>=85)
UNION ALL
(SELECT
    uid,
    "activity2" AS activity
FROM exam_record B
LEFT JOIN examination_info A
ON A.exam_id=B.exam_id
WHERE 
    TIMESTAMPDIFF(SECOND,start_time,submit_time)<=duration*30
    AND
    score>=80
    AND
    difficulty="hard"
GROUP BY uid)
ORDER BY uid

三、连接查询

1、满足条件的用户的试卷完成数和题目练习数

 

 自己的写法【错误】

SELECT
    A.uid,
    COUNT(DISTINCT C.uid) AS exam_cnt,
    COUNT(DISTINCT D.uid) AS question_cnt
FROM user_info A
JOIN examination_info B
JOIN exam_record C
JOIN practice_record D
ON 
    A.uid=C.uid
    AND
    A.uid=D.uid
    AND
    B.exam_id=C.exam_id
WHERE 
    level=7
    AND
    tag="SQL"
    AND
    difficulty="hard"
    AND
    (YEAR(C.submit_time)=2021
    OR
    YEAR(D.submit_time)=2021)
GROUP BY A.uid
HAVING -- 子句不能出现year
    AVG(C.score)>80
ORDER BY exam_cnt ASC,question_cnt DESC

答案：

-- 先分别写出2021年分组后的试卷完成情况和题目练习情况
-- 再查询出高难度SQL试卷得分平均值大于80并且是7级的红名大佬
select 
    uid,
    exam_cnt,
    if(question_cnt is null, 0, question_cnt)
from
(select 
    uid,
    count(submit_time) as exam_cnt
from exam_record
where YEAR(submit_time) = 2021
group by uid) t
 
left join
 
(select
    uid,
    count(submit_time) as question_cnt
from practice_record
where YEAR(submit_time) = 2021
group by uid) t2 using(uid)
 
where uid in
(
select
    uid
from exam_record
join examination_info using(exam_id)
join user_info using(uid)
where tag = 'SQL' and difficulty = 'hard' and `level` = 7
group by uid
having avg(score) >= 80
)
order by exam_cnt asc, question_cnt desc

2、每个6/7级用户活跃情况

 

自己分步写的：

-- 查询出6/7级用户
SELECT 
    uid
FROM user_info
WHERE 
    level=6
    OR
    level=7

-- 查询总活跃月份数
SELECT 
    uid,
    COUNT(DISTINCT act_month) AS act_month_total
FROM 
(SELECT 
    uid,
    DATE_FORMAT(start_time,"%Y%m") AS act_month
FROM exam_record
UNION ALL
SELECT 
    uid,
    DATE_FORMAT(submit_time,"%Y%m") AS act_month
FROM practice_record) t1
GROUP BY uid

-- 查询2021年活跃天数（UNION？）
SELECT 
    uid,
    COUNT(DISTINCT act_days) AS act_days_2021
FROM 
((SELECT 
    uid,
    DATE(start_time) AS act_days
FROM exam_record
WHERE 
    YEAR(start_time)=2021)
UNION ALL
(SELECT 
    uid,
    DATE(submit_time) AS act_days
FROM practice_record
WHERE 
    YEAR(submit_time)=2021)) t1
GROUP BY uid
        
-- 查询试卷作答活跃天数
SELECT 
    uid,
    COUNT(DISTINCT DATE(start_time)) AS act_month_total
FROM exam_record
WHERE YEAR(start_time)=2021
GROUP BY uid 

-- 2021年答题活跃天数
SELECT 
    uid,
    COUNT(DISTINCT DATE(submit_time)) AS act_month_total
FROM practice_record
WHERE YEAR(submit_time)=2021
GROUP BY uid

答案：

select
    ui.uid,
    count(distinct left(s,6)) as act_month_total,
    count(distinct if(left(s,4)='2021',right(s,4),null)) as act_days_2021,
    count(distinct if(left(s,4)='2021' and tag='e',right(s,4),null)) as act_days_2021_exam,
    count(distinct if(left(s,4)='2021' and tag='p',right(s,4),null)) as act_days_2021_question
from (
        select uid,DATE_FORMAT(submit_time,'%Y%m%d') as s,'p' tag from practice_record pr
        union all
        SELECT uid,DATE_FORMAT(start_time,'%Y%m%d') as s,'e' as tag from exam_record er
)mon
right join user_info ui
on ui.uid = mon.uid
where ui.level >5
group by uid
order by act_month_total DESC,act_days_2021 desc