牛客网SQL进阶127：月总刷题数和日均刷题数-编程知识

牛客网SQL进阶127：月总刷题数和日均刷题数

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月总刷题数和日均刷题数_牛客题霸_牛客网现有一张题目练习记录表practice_record，示例内容如下：。题目来自【牛客题霸】https://www.nowcoder.com/practice/f6b4770f453d4163acc419e3d19e6746?tpId=240

0 问题描述

基于练习记录表practice_record，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：

1 数据准备

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);

2 数据分析

方式一：union all 常规做法

-- 方式一：
selectDATE_FORMAT(submit_time,'%Y%m') as submit_month,count(1)as month_q_cnt,round(count(1)/ max(day(last_day(submit_time))) ,3) as avg_day_q_cnt
from practice_record
where year(submit_time) = '2021'group by DATE_FORMAT(submit_time,'%Y%m')
union all
select'2021汇总' as submit_month,count(1) as month_q_cnt,round(count(1) / 31 ,3) as avg_day_q_cnt
from practice_record 
where score is not null and year(submit_time) = '2021'
order by submit_month;

上述代码用到的函数：last_day()返回参数日期的最后一天，day(last_day())返回当月的天数

ps：这里最容易出错的点在于：每月天数的计算

(1) 计算每个月的天数可以用函数：day(last_day(datetime));
(2) 一年12月，每个月的天数： case when month(datetime) in (1,3,5,7,8,10,12) then 31 else 30 end
(3) 最容易出错的点在于 : group by DATE_FORMAT(submit_time,'%Y%m') 分组后，select后面只能跟：group by 分组字段、常量、以及 count()/ max()/min()/avg()/sum()等聚合函数；

由于 count(1) / max(day(last_day(submit_time)) 中分子count(1)用的是聚合函数，分母也必须用聚合函数，而函数day() 不是聚合函数，因此分母最终的逻辑为：max(day(last_day(submit_time)) 或min(day(last_day(submit_time))

方式二：with rollup

select coalesce(months,'2021汇总') as submit_month,count(question_id) as month_q_cnt,round(count(question_id)/max(days),3) as avg_day_cnt
from(select question_id,date_format(submit_time,'%Y%m') as months,day(last_day(submit_time)) as daysfrom practice_recordwhere year(submit_time)= '2021')  tmp1
group by months
with rollup;

上述代码拆解：

step1:利用date_format函数及day(last_day(submit_time)) 函数分别获取月份及当月的天数

select question_id,date_format(submit_time,'%Y%m') as months,day(last_day(submit_time)) as days
from practice_recordwhere year(submit_time)= '2021'

step2: 利用 group by with rollup 实现分组加和，利用ifnull/coalesce函数进行null值判断及补全：coalesce(months,'2021汇总') as submit_month

最终的代码如下：

select coalesce(months,'2021汇总') as submit_month,count(question_id) as month_q_cnt,round(count(question_id)/max(days),3) as avg_day_cnt
from(select question_id,date_format(submit_time,'%Y%m') as months,day(last_day(submit_time)) as daysfrom practice_recordwhere year(submit_time)= '2021')  tmp1
group by months
with rollup;

group by with rollup具体使用案例见文章：

MySQL ——group by子句使用with rollup-CSDN博客MySQL ——group by子句使用with rolluphttps://blog.csdn.net/SHWAITME/article/details/136078305?spm=1001.2014.3001.5502