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SQL16 得分不小于平均分的最低分
描述
示例数据 exam_record表（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分）：
| id | uid | exam_id | start_time | submit_time | score |
|---|---|---|---|---|---|
| 1 | 1001 | 9001 | 2020-01-02 09:01:01 | 2020-01-02 09:21:01 | 80 |
| 2 | 1002 | 9001 | 2021-09-05 19:01:01 | 2021-09-05 19:40:01 | 89 |
| 3 | 1002 | 9002 | 2021-09-02 12:01:01 | (NULL) | (NULL) |
| 4 | 1002 | 9003 | 2021-09-01 12:01:01 | (NULL) | (NULL) |
| 5 | 1002 | 9001 | 2021-02-02 19:01:01 | 2021-02-02 19:30:01 | 87 |
| 6 | 1002 | 9002 | 2021-05-05 18:01:01 | 2021-05-05 18:59:02 | 90 |
| 7 | 1003 | 9002 | 2021-02-06 12:01:01 | (NULL) | (NULL) |
| 8 | 1003 | 9003 | 2021-09-07 10:01:01 | 2021-09-07 10:31:01 | 86 |
| 9 | 1004 | 9003 | 2021-09-06 12:01:01 | (NULL) | (NULL) |
examination_info表（exam_id试卷ID, tag试卷类别, difficulty试卷难度, duration考试时长, release_time发布时间）
| id | exam_id | tag | difficulty | duration | release_time |
|---|---|---|---|---|---|
| 1 | 9001 | SQL | hard | 60 | 2020-01-01 10:00:00 |
| 2 | 9002 | SQL | easy | 60 | 2020-02-01 10:00:00 |
| 3 | 9003 | 算法 | medium | 80 | 2020-08-02 10:00:00 |
示例输出数据：
| min_score_over_avg |
|---|
| 87 |
解释：试卷9001和9002为SQL类别，作答这两份试卷的得分有[80,89,87,90]，平均分为86.5，不小于平均分的最小分数为87
问题
请从试卷作答记录表中找到SQL试卷得分不小于该类试卷平均得分的用户最低得分。
示例
输入：drop table if exists examination_info;
CREATE TABLE examination_info (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
exam_id int UNIQUE NOT NULL COMMENT '试卷ID',
tag varchar(32) COMMENT '类别标签',
difficulty varchar(8) COMMENT '难度',
duration int NOT NULL COMMENT '时长',
release_time datetime COMMENT '发布时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
drop table if exists exam_record;
CREATE TABLE exam_record (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid int NOT NULL COMMENT '用户ID',
exam_id int NOT NULL COMMENT '试卷ID',
start_time datetime NOT NULL COMMENT '开始时间',
submit_time datetime COMMENT '提交时间',
score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES
(9001, 'SQL', 'hard', 60, '2020-01-01 10:00:00'),
(9002, 'SQL', 'easy', 60, '2020-02-01 10:00:00'),
(9003, '算法', 'medium', 80, '2020-08-02 10:00:00');
INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2020-01-02 09:01:01', '2020-01-02 09:21:01', 80),
(1002, 9001, '2021-09-05 19:01:01', '2021-09-05 19:40:01', 89),
(1002, 9002, '2021-09-02 12:01:01', null, null),
(1002, 9003, '2021-09-01 12:01:01', null, null),
(1002, 9001, '2021-02-02 19:01:01', '2021-02-02 19:30:01', 87),
(1002, 9002, '2021-05-05 18:01:01', '2021-05-05 18:59:02', 90),
(1003, 9002, '2021-02-06 12:01:01', null, null),
(1003, 9003, '2021-09-07 10:01:01', '2021-09-07 10:31:01', 86),
(1004, 9003, '2021-09-06 12:01:01', null, null);
输出：87
解答
要计算得分的平均值，需要用到的知识点是：AVG()，此外，要从连接后的表中找到类别为 SQL 的事件的分数，需要用到条件筛查语句 SELECT ... FROM ... WHERE...。然后注意到试卷类别和得分信息处于两个不同的表，因此需要将两个表进行连接，此时需要用到连接语句 JOIN...ON...。
再然后，要找出不小于平均分的最低分，则需要用到聚合函数 MIN()。
SELECT MIN(exam_record.score) AS min_score_over_avg FROM exam_record
JOIN examination_info ON exam_record.exam_id = examination_info.exam_id
WHERE examination_info.tag = "SQL"
AND score >=
(SELECT AVG(exam_record.score) FROM exam_record JOIN examination_info ON exam_record.exam_id = examination_info.exam_id WHERE tag = "SQL")
SQL17 平均活跃天数和月活人数
描述
用户在牛客试卷作答区作答记录存储在表exam_record中，内容如下：
exam_record表（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分）
| id | uid | exam_id | start_time | submit_time | score |
|---|---|---|---|---|---|
| 1 | 1001 | 9001 | 2021-07-02 09:01:01 | 2021-07-02 09:21:01 | 80 |
| 2 | 1002 | 9001 | 2021-09-05 19:01:01 | 2021-09-05 19:40:01 | 81 |
| 3 | 1002 | 9002 | 2021-09-02 12:01:01 | (NULL) | (NULL) |
| 4 | 1002 | 9003 | 2021-09-01 12:01:01 | (NULL) | (NULL) |
| 5 | 1002 | 9001 | 2021-07-02 19:01:01 | 2021-07-02 19:30:01 | 82 |
| 6 | 1002 | 9002 | 2021-07-05 18:01:01 | 2021-07-05 18:59:02 | 90 |
| 7 | 1003 | 9002 | 2021-07-06 12:01:01 | (NULL) | (NULL) |
| 8 | 1003 | 9003 | 2021-09-07 10:01:01 | 2021-09-07 10:31:01 | 86 |
| 9 | 1004 | 9003 | 2021-09-06 12:01:01 | (NULL) | (NULL) |
| 10 | 1002 | 9003 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 81 |
| 11 | 1005 | 9001 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 88 |
| 12 | 1006 | 9002 | 2021-09-02 12:11:01 | 2021-09-02 12:31:01 | 89 |
| 13 | 1007 | 9002 |
问题
请计算2021年每个月里试卷作答区用户平均月活跃天数avg_active_days和月度活跃人数mau，上面数据的示例输出如下：
| month | avg_active_days | mau |
|---|---|---|
| 202107 | 1.50 | 2 |
| 202109 | 1.25 | 4 |
解释：2021年7月有2人活跃，共活跃了3天（1001活跃1天，1002活跃2天），平均活跃天数1.5；2021年9月有4人活跃，共活跃了5天，平均活跃天数1.25，结果保留2位小数。
注：此处活跃指有交卷行为。
示例
输入：
drop table if exists exam_record;
CREATE TABLE exam_record (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid int NOT NULL COMMENT '用户ID',
exam_id int NOT NULL COMMENT '试卷ID',
start_time datetime NOT NULL COMMENT '开始时间',
submit_time datetime COMMENT '提交时间',
score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-07-02 09:01:01', '2021-07-02 09:21:01', 80),
(1002, 9001, '2021-09-05 19:01:01', '2021-09-05 19:40:01', 81),
(1002, 9002, '2021-09-02 12:01:01', null, null),
(1002, 9003, '2021-09-01 12:01:01', null, null),
(1002, 9001, '2021-07-02 19:01:01', '2021-07-02 19:30:01', 82),
(1002, 9002, '2021-07-05 18:01:01', '2021-07-05 18:59:02', 90),
(1003, 9002, '2021-07-06 12:01:01', null, null),
(1003, 9003, '2021-09-07 10:01:01', '2021-09-07 10:31:01', 86),
(1004, 9003, '2021-09-06 12:01:01', null, null),
(1002, 9003, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 81),
(1005, 9001, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 88),
(1006, 9002, '2021-09-02 12:11:01', '2021-09-02 12:31:01', 89),
(1007, 9002, '2020-09-02 12:11:01', '2020-09-02 12:31:01', 89);
输出：
202107|1.50|2
202109|1.25|4
解答
按照月份分组，则需要用到 GROUP BY 知识点，同时要注意，因为是按月份分组，所以需要用到函数 DATE_FORMAT() 来匹配月份。同时需要筛选出活跃的用户，所以肯定少不了过滤语句 SELECT FROM ... FROM ...。最后，要进行统计，怎么能少的了 COUNT() 函数。
而再观察给出的结果，可以发现活跃天数都是保留的两位小数，所以此时还需要用到 ROUND()。
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 date_format(submit_time, '%Y%m')
SQL18 月总刷题数和日均刷题数
描述
现有一张题目练习记录表practice_record，示例内容如下：
| id | uid | question_id | submit_time | score |
|---|---|---|---|---|
| 1 | 1001 | 8001 | 2021-08-02 11:41:01 | 60 |
| 2 | 1002 | 8001 | 2021-09-02 19:30:01 | 50 |
| 3 | 1002 | 8001 | 2021-09-02 19:20:01 | 70 |
| 4 | 1002 | 8002 | 2021-09-02 19:38:01 | 70 |
| 5 | 1003 | 8002 | 2021-08-01 19:38:01 | 80 |
问题
请从中统计出2021年每个月里用户的月总刷题数month_q_cnt 和日均刷题数avg_day_q_cnt（按月份升序排序）以及该年的总体情况，示例数据输出如下：
| submit_month | month_q_cnt | avg_day_q_cnt |
|---|---|---|
| 202108 | 2 | 0.065 |
| 202109 | 3 | 0.100 |
| 2021汇总 | 5 | 0.161 |
解释：2021年8月共有2次刷题记录，日均刷题数为2/31=0.065（保留3位小数）；2021年9月共有3次刷题记录，日均刷题数为3/30=0.100；2021年共有5次刷题记录（年度汇总平均无实际意义，这里我们按照31天来算5/31=0.161）
示例
输入：
drop table if exists practice_record;
CREATE TABLE practice_record (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid int NOT NULL COMMENT '用户ID',
question_id int NOT NULL COMMENT '题目ID',
submit_time datetime COMMENT '提交时间',
score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
INSERT INTO practice_record(uid,question_id,submit_time,score) VALUES
(1001, 8001, '2021-08-02 11:41:01', 60),
(1002, 8001, '2021-09-02 19:30:01', 50),
(1002, 8001, '2021-09-02 19:20:01', 70),
(1002, 8002, '2021-09-02 19:38:01', 70),
(1003, 8002, '2021-08-01 19:38:01', 80);
输出：
202108|2|0.065
202109|3|0.100
2021汇总|5|0.161
解答
SELECT
COALESCE(DATE_FORMAT(submit_time,"%Y%m"),'2021汇总') submit_month,
COUNT(submit_time) month_q_cnt,
round(COUNT(submit_time) / MAX(DAY(last_day(submit_time))),3)avg_day_q_cnt
FROM
practice_record
WHERE
year(submit_time) = '2021'
GROUP BY
DATE_FORMAT(submit_time,"%Y%m") WITH ROLLUP;
SQL19 未完成试卷数大于1的有效用户
描述
现有试卷作答记录表exam_record（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分），示例数据如下：
| id | uid | exam_id | start_time | submit_time | score |
|---|---|---|---|---|---|
| 1 | 1001 | 9001 | 2021-07-02 09:01:01 | 2021-07-02 09:21:01 | 80 |
| 2 | 1002 | 9001 | 2021-09-05 19:01:01 | 2021-09-05 19:40:01 | 81 |
| 3 | 1002 | 9002 | 2021-09-02 12:01:01 | (NULL) | (NULL) |
| 4 | 1002 | 9003 | 2021-09-01 12:01:01 | (NULL) | (NULL) |
| 5 | 1002 | 9001 | 2021-07-02 19:01:01 | 2021-07-02 19:30:01 | 82 |
| 6 | 1002 | 9002 | 2021-07-05 18:01:01 | 2021-07-05 18:59:02 | 90 |
| 7 | 1003 | 9002 | 2021-07-06 12:01:01 | (NULL) | (NULL) |
| 8 | 1003 | 9003 | 2021-09-07 10:01:01 | 2021-09-07 10:31:01 | 86 |
| 9 | 1004 | 9003 | 2021-09-06 12:01:01 | (NULL) | (NULL) |
| 10 | 1002 | 9003 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 81 |
| 11 | 1005 | 9001 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 88 |
| 12 | 1005 | 9002 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 88 |
| 13 | 1006 | 9002 | 2021-09-02 12:11:01 | 2021-09-02 12:31:01 | 89 |
还有一张试卷信息表examination_info（exam_id试卷ID, tag试卷类别, difficulty试卷难度, duration考试时长, release_time发布时间），示例数据如下：
| id | exam_id | tag | difficulty | duration | release_time |
|---|---|---|---|---|---|
| 1 | 9001 | SQL | hard | 60 | 2020-01-01 10:00:00 |
| 2 | 9002 | SQL | easy | 60 | 2020-02-01 10:00:00 |
| 3 | 9003 | 算法 | medium | 80 | 2020-08-02 10:00:00 |
问题
请统计2021年每个未完成试卷作答数大于1的有效用户的数据（有效用户指完成试卷作答数至少为1且未完成数小于5），输出用户ID、未完成试卷作答数、完成试卷作答数、作答过的试卷tag集合，按未完成试卷数量由多到少排序。示例数据的输出结果如下：
| uid | incomplete_cnt | complete_cnt | detail |
|---|---|---|---|
| 1002 | 2 | 4 | 2021-09-01:算法;2021-07-02:SQL;2021-09-02:SQL;2021-09-05:SQL;2021-07-05:SQL |
解释：2021年的作答记录中，除了1004，其他用户均满足有效用户定义，但只有1002未完成试卷数大于1，因此只输出1002，detail中是1002作答过的试卷{日期:tag}集合，日期和tag间用:连接，多元素间用;连接。
示例
输入：
drop table if exists examination_info;
CREATE TABLE examination_info (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
exam_id int UNIQUE NOT NULL COMMENT '试卷ID',
tag varchar(32) COMMENT '类别标签',
difficulty varchar(8) COMMENT '难度',
duration int NOT NULL COMMENT '时长',
release_time datetime COMMENT '发布时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
drop table if exists exam_record;
CREATE TABLE exam_record (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid int NOT NULL COMMENT '用户ID',
exam_id int NOT NULL COMMENT '试卷ID',
start_time datetime NOT NULL COMMENT '开始时间',
submit_time datetime COMMENT '提交时间',
score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES
(9001, 'SQL', 'hard', 60, '2020-01-01 10:00:00'),
(9002, 'SQL', 'easy', 60, '2020-02-01 10:00:00'),
(9003, '算法', 'medium', 80, '2020-08-02 10:00:00');
INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-07-02 09:01:01', '2021-07-02 09:21:01', 80),
(1002, 9001, '2021-09-05 19:01:01', '2021-09-05 19:40:01', 81),
(1002, 9002, '2021-09-02 12:01:01', null, null),
(1002, 9003, '2021-09-01 12:01:01', null, null),
(1002, 9001, '2021-07-02 19:01:01', '2021-07-02 19:30:01', 82),
(1002, 9002, '2021-07-05 18:01:01', '2021-07-05 18:59:02', 90),
(1003, 9002, '2021-07-06 12:01:01', null, null),
(1003, 9003, '2021-09-07 10:01:01', '2021-09-07 10:31:01', 86),
(1004, 9003, '2021-09-06 12:01:01', null, null),
(1002, 9003, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 81),
(1005, 9001, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 88),
(1005, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 88),
(1006, 9002, '2021-09-02 12:11:01', '2021-09-02 12:31:01', 89);
输出：
1002|2|4|2021-07-02:SQL;2021-07-05:SQL;2021-09-01:算法;2021-09-02:SQL;2021-09-05:SQL
解答
SELECT uid, COUNT(incomplete) incomplete_cnt,
COUNT(complete) complete_cnt,
GROUP_CONCAT(DISTINCT concat_ws(':', DATE(start_time), tag) SEPARATOR ';') detail
FROM (
SELECT uid, tag, start_time,
IF(submit_time IS null, 1, null) AS incomplete,
IF(submit_time IS null, null, 1) AS complete
FROM exam_record
JOIN examination_info USING(exam_id)
WHERE year(start_time)=2021
) AS exam_complete_rec
GROUP BY uid
HAVING complete_cnt >= 1 AND incomplete_cnt BETWEEN 2 AND 4
ORDER BY incomplete_cnt DESC;
SQL20 月均完成试卷数不小于3的用户爱作答的类别
描述
现有试卷作答记录表exam_record（uid：用户ID, exam_id：试卷ID, start_time：开始作答时间, submit_time：交卷时间，没提交的话为NULL, score：得分），示例数据如下：
| id | uid | exam_id | start_time | submit_time | score |
|---|---|---|---|---|---|
| 1 | 1001 | 9001 | 2021-07-02 09:01:01 | (NULL) | (NULL) |
| 2 | 1002 | 9003 | 2021-09-01 12:01:01 | 2021-09-01 12:21:01 | 60 |
| 3 | 1002 | 9002 | 2021-09-02 12:01:01 | 2021-09-02 12:31:01 | 70 |
| 4 | 1002 | 9001 | 2021-09-05 19:01:01 | 2021-09-05 19:40:01 | 81 |
| 5 | 1002 | 9002 | 2021-07-06 12:01:01 | (NULL) | (NULL) |
| 6 | 1003 | 9003 | 2021-09-07 10:01:01 | 2021-09-07 10:31:01 | 86 |
| 7 | 1003 | 9003 | 2021-09-08 12:01:01 | 2021-09-08 12:11:01 | 40 |
| 8 | 1003 | 9001 | 2021-09-08 13:01:01 | (NULL) | (NULL) |
| 9 | 1003 | 9002 | 2021-09-08 14:01:01 | (NULL) | (NULL) |
| 10 | 1003 | 9003 | 2021-09-08 15:01:01 | (NULL) | (NULL) |
| 11 | 1005 | 9001 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 88 |
| 12 | 1005 | 9002 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 88 |
| 13 | 1005 | 9002 | 2021-09-02 12:11:01 | 2021-09-02 12:31:01 | 89 |
试卷信息表examination_info（exam_id：试卷ID, tag：试卷类别, difficulty：试卷难度, duration：考试时长, release_time：发布时间），示例数据如下：
| id | exam_id | tag | difficulty | duration | release_time |
|---|---|---|---|---|---|
| 1 | 9001 | SQL | hard | 60 | 2020-01-01 10:00:00 |
| 2 | 9002 | C++ | easy | 60 | 2020-02-01 10:00:00 |
| 3 | 9003 | 算法 | medium | 80 | 2020-08-02 10:00:00 |
问题
请从表中统计出 “当月均完成试卷数”不小于3的用户们爱作答的类别及作答次数，按次数降序输出，示例输出如下：
| tag | tag_cnt |
|---|---|
| C++ | 4 |
| SQL | 2 |
| 算法 | 1 |
解释：用户1002和1005在2021年09月的完成试卷数目均为3，其他用户均小于3；然后用户1002和1005作答过的试卷tag分布结果按作答次数降序排序依次为C++、SQL、算法。
示例
输入：
drop table if exists examination_info,exam_record;
CREATE TABLE examination_info (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
exam_id int UNIQUE NOT NULL COMMENT '试卷ID',
tag varchar(32) COMMENT '类别标签',
difficulty varchar(8) COMMENT '难度',
duration int NOT NULL COMMENT '时长',
release_time datetime COMMENT '发布时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
CREATE TABLE exam_record (
id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid int NOT NULL COMMENT '用户ID',
exam_id int NOT NULL COMMENT '试卷ID',
start_time datetime NOT NULL COMMENT '开始时间',
submit_time datetime COMMENT '提交时间',
score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;
INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES
(9001, 'SQL', 'hard', 60, '2020-01-01 10:00:00'),
(9002, 'C++', 'easy', 60, '2020-02-01 10:00:00'),
(9003, '算法', 'medium', 80, '2020-08-02 10:00:00');
INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-07-02 09:01:01', null, null),
(1002, 9003, '2021-09-01 12:01:01', '2021-09-01 12:21:01', 60),
(1002, 9002, '2021-09-02 12:01:01', '2021-09-02 12:31:01', 70),
(1002, 9001, '2021-09-05 19:01:01', '2021-09-05 19:40:01', 81),
(1002, 9002, '2021-07-06 12:01:01', null, null),
(1003, 9003, '2021-09-07 10:01:01', '2021-09-07 10:31:01', 86),
(1003, 9003, '2021-09-08 12:01:01', '2021-09-08 12:11:01', 40),
(1003, 9001, '2021-09-08 13:01:01', null, null),
(1003, 9002, '2021-09-08 14:01:01', null, null),
(1003, 9003, '2021-09-08 15:01:01', null, null),
(1005, 9001, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 88),
(1005, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 88),
(1005, 9002, '2021-09-02 12:11:01', '2021-09-02 12:31:01', 89);
输出：
C++|4
SQL|2
算法|1
解答
根据题意可知，本质主要是统计月均完成试卷数不少于 3 的用户所喜爱的类别和作答词数，然后按次数降序输出，将题中的知识点进行分解，就可以大致得到以下小知识点：
-
首先要进行筛选工作，那么肯定少不了
WHERE。 -
其次，要统计完成次数，那么需要用到分组语句
GROUP BY，然后要统计用户完成试卷的总数，需要用到统计某一列的聚合函数COUNT()，而要统计完成试卷的月份数，此时需要用到DATE_FORMAT()。 -
要将试卷作答记录和试卷信息两个表进行关联，需要用到
JOIN知识点。 -
最后则是降序输出，需要用到知识点
ORDER BY [列名] DESC。
SELECT tag, COUNT(tag) AS tag_cnt
FROM exam_record
JOIN examination_info USING(exam_id)
WHERE uid in
(SELECT uid FROM exam_record WHERE submit_time IS NOT NULL GROUP BY uid HAVING COUNT(exam_id) / COUNT(DISTINCT DATE_FORMAT(start_time, "%Y%m")) >= 3)
GROUP BY tag
ORDER BY
tag_cnt
DESC;
