[大数据]SQL 进阶挑战（21 - 25）

SQL21 试卷发布当天作答人数和平均分

描述

现有用户信息表user_info（uid用户ID，nick_name昵称, achievement成就值, level等级, job职业方向, register_time注册时间），示例数据如下：

释义：用户1001昵称为牛客1号，成就值为3100，用户等级是7级，职业方向为算法，注册时间2020-01-01 10:00:00

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：

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：

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：

问题

请计算每张SQL类别试卷发布后，当天5级以上的用户作答的人数uv和平均分avg_score，按人数降序，相同人数的按平均分升序，示例数据结果输出如下：

解释：只有一张SQL类别的试卷，试卷ID为9001，发布当天（2021-09-01）有1001、1002、1003、1005作答过，但是1003是5级用户，其他3位为5级以上，他们三的得分有[70,80,85,90]，平均分为81.3（保留1位小数）。

示例

输入：
drop table if exists examination_info,user_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 user_info (
    id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    uid int UNIQUE NOT NULL COMMENT '用户ID',
    `nick_name` varchar(64) COMMENT '昵称',
    achievement int COMMENT '成就值',
    level int COMMENT '用户等级',
    job varchar(32) COMMENT '职业方向',
    register_time datetime COMMENT '注册时间'
)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 user_info(uid,`nick_name`,achievement,level,job,register_time) VALUES
  (1001, '牛客1号', 3100, 7, '算法', '2020-01-01 10:00:00'),
  (1002, '牛客2号', 2100, 6, '算法', '2020-01-01 10:00:00'),
  (1003, '牛客3号', 1500, 5, '算法', '2020-01-01 10:00:00'),
  (1004, '牛客4号', 1100, 4, '算法', '2020-01-01 10:00:00'),
  (1005, '牛客5号', 1600, 6, 'C++', '2020-01-01 10:00:00'),
  (1006, '牛客6号', 3000, 6, 'C++', '2020-01-01 10:00:00');

INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES
  (9001, 'SQL', 'hard', 60, '2021-09-01 06:00:00'),
  (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, '2021-09-01 09:01:01', '2021-09-01 09:41:01', 70),
(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, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 80),
(1002, 9003, '2021-08-01 12:01:01', '2021-08-01 12:21:01', 60),
(1002, 9002, '2021-08-02 12:01:01', '2021-08-02 12:31:01', 70),
(1002, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 85),
(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, '2021-09-01 13:01:01', '2021-09-01 13:41:01', 70),
(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', 90),
(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：
9001|3|81.3

解答

对问题进行分解：

1. 要筛选 SQL 类别试卷，所以需要用到条件筛查语句：WHERE tag = "SQL"。
2. 为了获取试卷 ID 和发布日期，所以需要查询语句：SELECT exam_id, DATE(release_time)。
3. 筛选当日作答记录：WHERE (exam_id, DATE(start_time)) IN ...。
4. 5 级以上用户：SELECT uid FROM user_info WHERE level > 5。
5. 按试卷 ID 分类：GROUP BY exam_id。
6. 清点作答人数：COUNT(DISTINCT uid) AS uv。
7. 计算平均分：ROUND(AVG(score), 1) AS avg_score。

最终进行组合，得到以下结果：

SELECT
    exam_id,
    COUNT(DISTINCT uid) AS uv,
    ROUND(avg(score), 1) AS avg_score
FROM exam_record 
WHERE (exam_id, DATE(start_time)) IN (
    SELECT exam_id, DATE(release_time)
    FROM examination_info WHERE tag = "SQL") 
AND uid IN ( SELECT uid FROM user_info WHERE `level` > 5 )
GROUP BY exam_id
ORDER BY uv DESC, avg_score ASC;

SQL22 作答试卷得分大于过80的人的用户等级分布

描述

现有用户信息表user_info（uid用户ID，nick_name昵称, achievement成就值, level等级, job职业方向, register_time注册时间）：

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）：

试卷作答信息表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）：

问题

统计作答SQL类别的试卷得分大于过80的人的用户等级分布，按数量降序排序（保证数量都不同）。示例数据结果输出如下：

解释：9001为SQL类试卷，作答该试卷大于80分的人有1002、1003、1005共3人，6级两人，5级一人。

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 user_info;
CREATE TABLE user_info (
    id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    uid int UNIQUE NOT NULL COMMENT '用户ID',
    `nick_name` varchar(64) COMMENT '昵称',
    achievement int COMMENT '成就值',
    level int COMMENT '用户等级',
    job varchar(32) COMMENT '职业方向',
    register_time datetime COMMENT '注册时间'
)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 user_info(uid,`nick_name`,achievement,level,job,register_time) VALUES
  (1001, '牛客1号', 3100, 7, '算法', '2020-01-01 10:00:00'),
  (1002, '牛客2号', 2100, 6, '算法', '2020-01-01 10:00:00'),
  (1003, '牛客3号', 1500, 5, '算法', '2020-01-01 10:00:00'),
  (1004, '牛客4号', 1100, 4, '算法', '2020-01-01 10:00:00'),
  (1005, '牛客5号', 1600, 6, 'C++', '2020-01-01 10:00:00'),
  (1006, '牛客6号', 3000, 6, 'C++', '2020-01-01 10:00:00');

INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES
  (9001, 'SQL', 'hard', 60, '2021-09-01 06:00:00'),
  (9002, 'C++', 'easy', 60, '2021-09-01 06:00:00'),
  (9003, '算法', 'medium', 80, '2021-09-01 10:00:00');

INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-09-01 09:01:01', '2021-09-01 09:41:01', 79),
(1002, 9003, '2021-09-01 12:01:01', '2021-09-01 12:21:01', 60),
(1002, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 70),
(1002, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 80),
(1002, 9003, '2021-08-01 12:01:01', '2021-08-01 12:21:01', 60),
(1002, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 70),
(1002, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 85),
(1002, 9002, '2021-09-01 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, '2021-09-01 13:01:01', '2021-09-01 13:41:01', 81),
(1003, 9002, '2021-09-01 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', 90),
(1005, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 88),
(1006, 9002, '2021-09-01 12:11:01', '2021-09-01 12:31:01', 89);
输出：
6|2
5|1

解答

将问题进行拆分，分别来看看用到了哪些知识点：

1. 要按照等级分组，所以肯定会用到 GROUP BY [列名]。
2. 而要对类别、得分等信息组合到一起，肯定需要将多个表进行连接，此时就需要用到 JOIN ... ON...。
3. 同时，需要筛选出类别和得分满足一定条件的记录，所以需要用到 WHERE [条件]。
4. 最后则是排序，用到的知识点是 ORDER BY [列名] [ASC|DESC]

SELECT level,
       COUNT(DISTINCT user_info.uid) AS level_cnt
FROM exam_record JOIN examination_info 
ON exam_record.exam_id = examination_info.exam_id
JOIN user_info 
ON exam_record.uid = user_info.uid
WHERE tag = 'SQL'
AND score > 80
GROUP BY level 
ORDER BY level_cnt DESC, level DESC

SQL23 每个题目和每份试卷被作答的人数和次数

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）：

题目练习表practice_record（uid用户ID, question_id题目ID, submit_time提交时间, score得分）：

问题

请统计每个题目和每份试卷被作答的人数和次数，分别按照"试卷"和"题目"的uv & pv降序显示，示例数据结果输出如下：

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;

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 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, 8001, '2021-08-02 19:38:01', 70),
(1003, 8001, '2021-08-02 19:48:01', 90),
(1003, 8002, '2021-08-01 19:38:01', 80);

INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-09-01 09:01:01', '2021-09-01 09:41:01', 81),
(1002, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 70),
(1002, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 80),
(1002, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 70),
(1004, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 85),
(1002, 9002, '2021-09-01 12:01:01', null, null);
输出：
9001|3|3
9002|1|3
8001|3|5
8002|2|2

解答

主要考察 UNION 操作符，用于将两个或多个 SELECT 语句的结果集进行合并，其语法如下：

SELECT column_name(s) FROM table1
UNION
SELECT column_name(s) FROM table2;

需要注意的是 UNION 内部的每个 SELECT 语句都必须拥有相同数量的列，而且列的数据类型也应该相似，同时需要两个 SELECT 语句中的列的顺序也相同。

SELECT * FROM
(SELECT exam_id AS tid, COUNT(DISTINCT exam_record.uid) uv,
COUNT(*) pv FROM exam_record 
GROUP BY exam_id
ORDER BY uv DESC, pv DESC) t1
UNION 
SELECT * FROM
(SELECT question_id AS tid, COUNT(DISTINCT practice_record.uid) uv,
COUNT(*) pv FROM practice_record 
GROUP BY question_id
ORDER BY uv DESC, pv DESC) t2;

SQL24 分别满足两个活动的人

描述

为了促进更多用户在牛客平台学习和刷题进步，我们会经常给一些既活跃又表现不错的用户发放福利。假使以前我们有两拨运营活动，分别给每次试卷得分都能到85分的人（activity1）、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（activity2）发了福利券。

现有试卷信息表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）：

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）：

示例数据输出结果：

解释：用户1001最小分数81不满足活动1，但29分59秒完成了60分钟长的试卷得分81，满足活动2；1003最小分数86满足活动1，完成时长都大于试卷时长的一半，不满足活动2；用户1004刚好用了一半时间（30分钟整）完成了试卷得分85，满足活动1和活动2。

问题

现在，需要你一次性将这两个活动满足的人筛选出来，交给运营同学。请写出一个SQL实现：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。

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, '2021-09-01 06:00:00'),
  (9002, 'C++', 'hard', 60, '2021-09-01 06:00:00'),
  (9003, '算法', 'medium', 80, '2021-09-01 10:00:00');

INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-09-01 09:01:01', '2021-09-01 09:31:00', 81),
(1002, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 70),
(1003, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 86),
(1003, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:51', 89),
(1004, 9001, '2021-09-01 19:01:01', '2021-09-01 19:30:01', 85);
输出：
1001|activity2
1003|activity1
1004|activity1
1004|activity2

解答

和上一题类似，这里有两个条件，而且查询字段都相同，所以需要将两次查询后的结果相联合，此时急需要用到 UNION，但不同的是，此题不需要去重，所以需要使用 UNION ALL。

首先来看第一个条件，查询 2021 年所有得分不小于 85 分的试卷。

SELECT
	uid,
	'activity1' AS 'activity' 
FROM
	exam_record 
WHERE YEAR ( start_time ) = 2021 
GROUP BY uid 
HAVING MIN( score ) >= 85 

再来看看第二个条件，分数大于 80，而且至少有一次用时只占规定的一半就完成了高难度试卷。

SELECT 
	DISTINCT t1.uid,
	'activity2' AS 'activity'
FROM
	exam_record AS t1
INNER JOIN examination_info AS t2 ON t1.exam_id = t2.exam_id 
WHERE TIMESTAMPDIFF( SECOND, t1.start_time, t1.submit_time ) <= t2.duration * 30 
AND t2.difficulty = 'hard' AND score > 80 AND YEAR ( t1.start_time ) = 2021 

最后将两个条件进行连接组合即可。

SELECT
	uid,
	'activity1' AS 'activity' 
FROM
	exam_record 
WHERE YEAR ( start_time ) = 2021 
GROUP BY uid 
HAVING MIN( score ) >= 85 
UNION ALL
SELECT 
	DISTINCT t1.uid,
	'activity2' AS 'activity'
FROM
	exam_record AS t1
INNER JOIN examination_info AS t2 ON t1.exam_id = t2.exam_id 
WHERE TIMESTAMPDIFF( SECOND, t1.start_time, t1.submit_time ) <= t2.duration * 30 
AND t2.difficulty = 'hard' AND score > 80 AND YEAR ( t1.start_time ) = 2021 
ORDER BY uid;

SQL25 满足条件的用户的试卷完成数和题目练习数

描述

现有用户信息表user_info（uid用户ID，nick_name昵称, achievement成就值, level等级, job职业方向, register_time注册时间）：

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）：

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）：

题目练习记录表practice_record（uid用户ID, question_id题目ID, submit_time提交时间, score得分）：

问题

请你找到高难度SQL试卷得分平均值大于80并且是7级的红名大佬，统计他们的2021年试卷总完成次数和题目总练习次数，只保留2021年有试卷完成记录的用户。结果按试卷完成数升序，按题目练习数降序。

示例数据输出如下：

解释：用户1001、1003、1004、1006满足高难度SQL试卷得分平均值大于80，但只有1001、1003是7级红名大佬；1001完成了1次试卷1001，练习了2次题目；1003完成了2次试卷9001、9002，未练习题目（因此计数为0）

示例

输入：
drop table if exists examination_info,user_info,exam_record,practice_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 user_info (
    id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    uid int UNIQUE NOT NULL COMMENT '用户ID',
    `nick_name` varchar(64) COMMENT '昵称',
    achievement int COMMENT '成就值',
    level int COMMENT '用户等级',
    job varchar(32) COMMENT '职业方向',
    register_time datetime COMMENT '注册时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;

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;

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 user_info(uid,`nick_name`,achievement,level,job,register_time) VALUES
  (1001, '牛客1号', 3100, 7, '算法', '2020-01-01 10:00:00'),
  (1002, '牛客2号', 2300, 7, '算法', '2020-01-01 10:00:00'),
  (1003, '牛客3号', 2500, 7, '算法', '2020-01-01 10:00:00'),
  (1004, '牛客4号', 1200, 5, '算法', '2020-01-01 10:00:00'),
  (1005, '牛客5号', 1600, 6, 'C++', '2020-01-01 10:00:00'),
  (1006, '牛客6号', 2000, 6, 'C++', '2020-01-01 10:00:00');

INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES
  (9001, 'SQL', 'hard', 60, '2021-09-01 06:00:00'),
  (9002, 'C++', 'hard', 60, '2021-09-01 06:00:00'),
  (9003, '算法', 'medium', 80, '2021-09-01 10:00:00');

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),
(1004, 8001, '2021-08-02 19:38:01', 70),
(1004, 8002, '2021-08-02 19:48:01', 90),
(1001, 8002, '2021-08-02 19:38:01', 70),
(1004, 8002, '2021-08-02 19:48:01', 90),
(1004, 8002, '2021-08-02 19:58:01', 94),
(1004, 8003, '2021-08-02 19:38:01', 70),
(1004, 8003, '2021-08-02 19:48:01', 90),
(1004, 8003, '2021-08-01 19:38:01', 80);

INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2021-09-01 09:01:01', '2021-09-01 09:31:00', 81),
(1002, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:01', 81),
(1003, 9001, '2021-09-01 19:01:01', '2021-09-01 19:40:01', 86),
(1003, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:51', 89),
(1004, 9001, '2021-09-01 19:01:01', '2021-09-01 19:30:01', 85),
(1005, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:02', 85),
(1006, 9003, '2021-09-07 10:01:01', '2021-09-07 10:21:01', 84),
(1006, 9001, '2021-09-07 10:01:01', '2021-09-07 10:21:01', 80);
输出：
1001|1|2
1003|2|0

解答

问题很复杂，所以选择一步步分解后再组合。

首先，我们将完成的试卷数和题目练习数目统计出来。

SELECT 
    uid,
    COUNT(submit_time) AS exam_cnt
from exam_record
WHERE YEAR(submit_time) = 2021
GROUP BY uid

SELECT
    uid,
    COUNT(submit_time) AS question_cnt
FROM practice_record
WHERE YEAR(submit_time) = 2021
GROUP BY uid

接着筛进行筛选，将 SQL 试卷且难度为高难度，得分平均分超过 80 分，用户是 7 级的用户找出来。

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

最后，将上面的语句进行组装，就可以得到最终的答案了！

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) table1
LEFT JOIN
(SELECT
    uid,
    COUNT(submit_time) AS question_cnt
FROM practice_record
WHERE YEAR(submit_time) = 2021
GROUP BY uid) table2 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