sql刷题

sql刷题"/>

sql刷题

计算用户的平均次日留存率

关键字和函数

关键词 distinct 用于返回唯一不同的值。

DATE_ADD() 函数向日期添加指定的时间间隔。

问题描述

题目：现在运营想要查看用户在某天刷题后第二天还会再来刷题的平均概率。请你取出相应数据。

根据示例，你的查询应返回以下结果：

示例

drop table if exists `user_profile`;
drop table if  exists `question_practice_detail`;
drop table if  exists `question_detail`;
CREATE TABLE `user_profile` (
`id` int NOT NULL,
`device_id` int NOT NULL,
`gender` varchar(14) NOT NULL,
`age` int ,
`university` varchar(32) NOT NULL,
`gpa` float,
`active_days_within_30` int ,
`question_cnt` int ,
`answer_cnt` int 
);
CREATE TABLE `question_practice_detail` (
`id` int NOT NULL,
`device_id` int NOT NULL,
`question_id`int NOT NULL,
`result` varchar(32) NOT NULL,
`date` date NOT NULL
);
CREATE TABLE `question_detail` (
`id` int NOT NULL,
`question_id`int NOT NULL,
`difficult_level` varchar(32) NOT NULL
);INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4,7,2,12);
INSERT INTO user_profile VALUES(2,3214,'male',null,'复旦大学',4.0,15,5,25);
INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2,12,3,30);
INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6,5,1,2);
INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8,20,15,70);
INSERT INTO user_profile VALUES(6,2131,'male',28,'山东大学',3.3,15,7,13);
INSERT INTO user_profile VALUES(7,4321,'male',28,'复旦大学',3.6,9,6,52);
INSERT INTO question_practice_detail VALUES(1,2138,111,'wrong','2021-05-03');
INSERT INTO question_practice_detail VALUES(2,3214,112,'wrong','2021-05-09');
INSERT INTO question_practice_detail VALUES(3,3214,113,'wrong','2021-06-15');
INSERT INTO question_practice_detail VALUES(4,6543,111,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(5,2315,115,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(6,2315,116,'right','2021-08-14');
INSERT INTO question_practice_detail VALUES(7,2315,117,'wrong','2021-08-15');
INSERT INTO question_practice_detail VALUES(8,3214,112,'wrong','2021-05-09');
INSERT INTO question_practice_detail VALUES(9,3214,113,'wrong','2021-08-15');
INSERT INTO question_practice_detail VALUES(10,6543,111,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(11,2315,115,'right','2021-08-13');
INSERT INTO question_practice_detail VALUES(12,2315,116,'right','2021-08-14');
INSERT INTO question_practice_detail VALUES(13,2315,117,'wrong','2021-08-15');
INSERT INTO question_practice_detail VALUES(14,3214,112,'wrong','2021-08-16');
INSERT INTO question_practice_detail VALUES(15,3214,113,'wrong','2021-08-18');
INSERT INTO question_practice_detail VALUES(16,6543,111,'right','2021-08-13');
INSERT INTO question_detail VALUES(1,111,'hard');
INSERT INTO question_detail VALUES(2,112,'medium');
INSERT INTO question_detail VALUES(3,113,'easy');
INSERT INTO question_detail VALUES(4,115,'easy');
INSERT INTO question_detail VALUES(5,116,'medium');
INSERT INTO question_detail VALUES(6,117,'easy');

解题

分析

关键词1第二天,如下，然后根据用户device_id来判断用户每个用户第二天有没有刷题。

date_add(date,interval 1 day)

关键词2平均概率,如下

通过关键词一，可以确定，当用户第二天打卡，则第二天会有数据，所以以第二天为分子，第一天为分母即可得到平均概率

实施

步骤一、用于返回每个用户每天唯一的数据

 select distinct device_id, date from question_practice_detail

步骤二、每个用户通过第一天数据和第二天数据做比较，然后使用第二天的数据做分子，第一天的数据做分子得到最后的结果。子查询一定要重命名。

select count(date2) / count(date1) as avg_ret
from (selectdistinct qpd.device_id,qpd.date as date1,uniq_id_date.date as date2from question_practice_detail as qpdleft join(select distinct device_id, datefrom question_practice_detail) as uniq_id_dateon qpd.device_id=uniq_id_date.device_idand date_add(qpd.date, interval 1 day)=uniq_id_date.date
) as id_last_next_date