【SQL进阶】【CASE/IF、COUNT/SUM、多条记录拼接为一个内容】Day03：聚合分组查询

〇、今日内容概述

一、聚合函数

1、SQL类别高难度试卷得分的截断平均值【去最高最低分求平均】

• 自己的想法
  
  SELECT tag, difficulty, ROUND((SUM(score)-MIN(score)-MAX(score))/(COUNT(score)-2),1) AS clip_avg_score FROM examination_info,exam_record WHERE examination_info.exam_id=exam_record.exam_id AND tag='hard' AND tag='SQL'

• 报错：
  
  Execution Error SQL_ERROR_INFO: "In aggregated query without GROUP BY, expression #2 of SELECT list contains nonaggregated column 'examination_info.difficulty'; this is incompatible with sql_mode=only_full_group_by"

• 正确做法
  
  SELECT tag, difficulty, ROUND((SUM(score)-MIN(score)-MAX(score))/(COUNT(score)-2),1) AS clip_avg_score FROM examination_info JOIN exam_record USING(exam_id) WHERE tag='SQL' AND difficulty='hard'
  
  或
  
  SELECT tag, difficulty, ROUND((SUM(score)-MIN(score)-MAX(score))/(COUNT(score)-2),1) AS clip_avg_score FROM examination_info,exam_record WHERE examination_info.exam_id=exam_record.exam_id AND difficulty='hard' AND tag='SQL'

2、统计作答次数

• 自己的想法
  
  SELECT COUNT(*) AS total_pv, SUM((CASE WHEN score IS NULL AND submit_time IS NULL THEN 0 ELSE 1 )) AS complete_pv, COUNT(DISTINCT exam_id) AS complete_exam_cnt FROM exam_record
• 报错
  
  Execution Error SQL_ERROR_INFO: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')) AS complete_pv,\n COUNT(DISTINCT exam_id) AS complete_exam_cnt\nFROM exam_re' at line 7"
• 正确做法
  
  SELECT COUNT(*) AS total_pv, SUM((CASE WHEN score IS NULL AND submit_time IS NULL THEN 0 ELSE 1 END )) AS complete_pv, COUNT(DISTINCT exam_id AND score IS NOT NULL AND submit_time IS NOT NULL) AS complete_exam_cnt FROM exam_record
• 原因：CASE ... WHEN ... THEN ... END
• 方法2：使用IF
  
  SELECT COUNT(*) AS total_pv, SUM((CASE WHEN score IS NULL AND submit_time IS NULL THEN 0 ELSE 1 END )) AS complete_pv, COUNT(DISTINCT IF(score IS NOT NULL, exam_id, NULL)) AS complete_exam_cnt FROM exam_record

3、得分不小于平均分的最低分

• 分组的可以在分组内使用join
  
  
  
  
• 自己的思路

点击查看代码

SELECT
    MIN(score) AS min_score_over_avg
FROM exam_record A
JOIN examination_info B
JOIN (SELECT exam_id,AVG(score) AS ex_score
      FROM exam_record
      GROUP BY exam_id) AVG_E
USING exam_id
WHERE
    score<ex_score
    AND
    tag='SQL'

* 正确答案

点击查看代码

SELECT
    MIN(score) AS min_score_over_avg
FROM exam_record er
JOIN examination_info ei
ON er.exam_id=ei.exam_id
WHERE
    tag='SQL'
    AND score>=
    (SELECT AVG(score)
    FROM exam_record er
    WHERE
        tag='SQL'
        AND
        er.exam_id=ei.exam_id
    GROUP BY er.exam_id)

* 方法2：使用over函数☆
# 二、分组函数
## 1、平均**活跃天数**和月活人数

• 自己写的

点击查看代码

SELECT
    DATE_FORMAT(submit_time,"%Y%m") AS month,
    ROUND(SUM(IF(submit_time IS NOT NULL,1,0))/COUNT(DISTINCT uid),2) AS avg_active_days,
    COUNT(DISTINCT uid) AS mau
FROM exam_record
WHERE
    submit_time IS NOT NULL
    AND
    YEAR(submit_time)='2021'
GROUP BY month

* 陷阱在于九月份有个用户同一天做了两种卷子，直接count统计的话活跃天数会多一天，即用户ID和做题日期submit_time要同时去重才能得出正确的活跃天数.
* 正确答案

点击查看代码

SELECT
    DATE_FORMAT(submit_time,"%Y%m") AS month,
    ROUND(COUNT(DISTINCT uid,DATE_FORMAT(submit_time,"%Y%m%d"))/COUNT(DISTINCT uid),2) AS avg_active_days,
    COUNT(DISTINCT uid) AS mau
FROM exam_record
WHERE
    submit_time IS NOT NULL
    AND
    YEAR(submit_time)='2021'
GROUP BY month

## 2、月总刷题数和日均刷题数【拼接未知数据使用UNION】

• 自己写的【错误】：

点击查看代码

SELECT
    DATE_FORMAT(submit_time,"%Y%m") AS submit_month,
    COUNT(submit_time) AS month_q_cnt,
    ROUND(COUNT(submit_time)/(
        CASE
            WHEN MONTH(submit_time)=1 THEN 31
            WHEN MONTH(submit_time)=2 THEN 28
            WHEN MONTH(submit_time)=3 THEN 31
            WHEN MONTH(submit_time)=4 THEN 30
            WHEN MONTH(submit_time)=5 THEN 31
            WHEN MONTH(submit_time)=6 THEN 30
            WHEN MONTH(submit_time)=7 THEN 31
            WHEN MONTH(submit_time)=8 THEN 31
            WHEN MONTH(submit_time)=9 THEN 30
            WHEN MONTH(submit_time)=10 THEN 31
            WHEN MONTH(submit_time)=11 THEN 30
            WHEN MONTH(submit_time)=12 THEN 31
        END
    ),3) AS avg_day_q_cnt
FROM practice_record
WHERE
    submit_time IS NOT NULL
    AND
    YEAR(submit_time)=2021
GROUP BY submit_month
ORDER BY submit_month ASC

• 正确答案

点击查看代码

SELECT
    DATE_FORMAT(submit_time,"%Y%m") submit_month,
    COUNT(submit_time) month_q_cnt,
    ROUND(COUNT(submit_time)/MAX(DAY(LAST_DAY(submit_time))),3) avg_day_q_cnt
    -- 使用max实现去重
FROM practice_record
WHERE YEAR(submit_time)=2021
GROUP BY submit_month
UNION ALL
SELECT
    "2021汇总" submit_month,
    COUNT(submit_time) month_q_cnt,
    ROUND(COUNT(submit_time)/31,3) avg_day_q_cnt
FROM practice_record
WHERE YEAR(submit_time)=2021
ORDER BY submit_month ASC

3、未完成试卷数大于1的有效用户

点击查看代码

SELECT
    uid,
    SUM(IF(er.submit_time IS NULL,1,0)) AS incomplete_cnt,
    -- COUNT(CASE WHEN er.submit_time IS NULL THEN er.start_time ELSE NULL END) AS incomplete_cnt,
    SUM(IF(er.submit_time IS NOT NULL,1,0)) AS complete_cnt,
    **GROUP_CONCAT(DISTINCT CONCAT_WS(':',DATE_FORMAT(er.start_time,"%Y-%m-%d"),ei.tag) SEPARATOR ';') **AS detail
FROM exam_record er
LEFT JOIN examination_info ei
ON er.exam_id=ei.exam_id
WHERE YEAR(er.start_time)=2021
GROUP BY er.uid
HAVING
    complete_cnt>=1
    AND
    incomplete_cnt<5
    AND
    incomplete_cnt>1
ORDER BY incomplete_cnt DESC