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第二快慢用时之差大于试卷时长一半的试卷_牛客题霸_牛客网现有试卷信息表examination_info（exam_id试卷ID, tag试卷类别,。题目来自【牛客题霸】
https://www.nowcoder.com/practice/b1e2864271c14b63b0df9fc08b559166?tpId=240
0 问题描述
- 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）
- 试卷信息表examination_info和试卷作答记录表exam_record, 找到第二快和第二慢用时之差大于试卷时长的一半的试卷信息，按试卷ID降序排序
1 数据准备
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, '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 09:51:01', 78),
(1001, 9002, '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 19:59:01', 86),
(1003, 9002, '2021-09-01 12:01:01', '2021-09-01 12:31:51', 89),
(1004, 9002, '2021-09-01 19:01:01', '2021-09-01 19:30:01', 85),
(1005, 9001, '2021-09-01 12:01:01', '2021-09-01 12:31:02', 85),
(1006, 9001, '2021-09-07 10:01:01', '2021-09-07 10:12:01', 84),
(1003, 9001, '2021-09-08 12:01:01', '2021-09-08 12:11:01', 40),
(1003, 9002, '2021-09-01 14:01:01', null, null),
(1005, 9001, '2021-09-01 14:01:01', null, null),
(1003, 9003, '2021-09-08 15:01:01', null, null);2 数据分析
完整的代码如下：
select distinct exam_id,
duration,
release_time
from(select exam_id,
duration,
release_time,
sum(case when rn1 =2 then difftime
when rn2 =2 then -difftime
else 0
end ) as sub
from(select
exam_id,
duration,
release_time,
difftime,
row_number() over(partition by exam_id order by difftime desc ) as rn1,
row_number() over(partition by exam_id order by difftime ) as rn2
from (select
er.exam_id,
ei.duration,
ei.release_time,
timestampdiff(minute,er.start_time,er.submit_time) as difftime
from exam_record er join examination_info ei
on er.exam_id = ei.exam_id
where submit_time is not null)tmp1
)tmp2
group by exam_id
)tmp3
where sub * 2 >= duration
order by exam_id desc;上述的解题步骤拆分
step1:求出各试卷的用时之差，并进行正序、逆序排序
step2:求出第二快和第二慢的用时之差，并和试卷规定时长（duration）进行比对step3:试卷ID降序排序
步骤代码
step1:
select
exam_id,
duration,
release_time,
difftime,
--进行正序、逆序排序
row_number() over(partition by exam_id order by difftime desc ) as rn1,
row_number() over(partition by exam_id order by difftime ) as rn2
from (select
er.exam_id,
ei.duration,
ei.release_time,
--step1:求出各试卷的用时之差timestampdiff，并进行正序、逆序排序
timestampdiff(minute,er.start_time,er.submit_time) as difftime
from exam_record er join examination_info ei
on er.exam_id = ei.exam_id
where submit_time is not null)tmp1;step2: 使用 case when进行赋值，当rn1 =2 时，代表是第二快的difftime(取正值)；当rn2 =2 时，代表是第二慢的difftime(需要取负值)； 外层再嵌套sum聚合函数，即得到第二快和第二慢的用时之差sub
select exam_id,
duration,
release_time,
sum(case when rn1 =2 then difftime
when rn2 =2 then -difftime
else 0
end ) as sub
from(select
exam_id,
duration,
release_time,
difftime,
row_number() over(partition by exam_id order by difftime desc ) as rn1,
row_number() over(partition by exam_id order by difftime ) as rn2
from (select
er.exam_id,
ei.duration,
ei.release_time,
timestampdiff(minute,er.start_time,er.submit_time) as difftime
from exam_record er join examination_info ei
on er.exam_id = ei.exam_id
where submit_time is not null)tmp1
)tmp2
group by exam_id;step3: sub和试卷规定时长（duration）进行比对，要求：sub * 2 >= duration
select distinct exam_id,
duration,
release_time
from(select exam_id,
duration,
release_time,
sum(case when rn1 =2 then difftime
when rn2 =2 then -difftime
else 0
end ) as sub
from(select
exam_id,
duration,
release_time,
difftime,
row_number() over(partition by exam_id order by difftime desc ) as rn1,
row_number() over(partition by exam_id order by difftime ) as rn2
from (select
er.exam_id,
ei.duration,
ei.release_time,
timestampdiff(minute,er.start_time,er.submit_time) as difftime
from exam_record er join examination_info ei
on er.exam_id = ei.exam_id
where submit_time is not null)tmp1
)tmp2
group by exam_id
)tmp3
where sub * 2 >= duration
order by exam_id desc;3 小结
上述案例用到的知识点：
（1）timestampdiff函数
timestampdiff： MySQL 中用来计算两个日期或时间之间的差值的函数；
语法：timestampdiff(unit, start_date, end_date)
参数说明：
unit：差值的单位，可以是
second（秒）、minute（分）、hour（小时）、day（天）、week（周）、month（月）、quarter（季度）或year（年）。
start_date：表示时间段的起始时间end_date：表示时间段的结束时间
（2）row_number() over(partition by ..order by ..desc)窗口函数
（3）sum +case when ：条件+聚合
