SQL:结构化查询语言(Structured Query Language)
留存率
现有一表Retention如下
id | name | register_time | logout_time |
---|
101 | Anna | 2018-01-01 00:00:01.000 | 2018-01-02 00:00:01.003 |
… | … | … | … |
register_time表示用户注册时间,logout_time表示用户注销时间。现要计算每一日的用户次日留存率(假如1号有5人注册,这5人有2人在2号注销,则1号当天的次日留存率为3/5=60%)
1、建表
- CREATE TABLE Retention
- (
- id int NOT NULL PRIMARY KEY,
- name nvarchar(50) NOT NULL,
- register_timedatetime NOT NULL,
- logout_timedatetime NOT NULL,
- );
|
2、插入数据
- INSERT INTO Retention
- VALUES (101, 'Anna', CONVERT(datetime,2018-01-01 00:00:01', 20), CONVERT(datetime,2018-01-01 00:00:01', 20));
- ...
|
3、更新数据
- UPDATE Retention
- SET logout_time = convert(datetime, 2018-01-02 00:00:02.003', 20)
- WHERE id=101;
|
4、计算留存率
当前表中数据如下:

SQL代码如下:
- SELECT new.register_day AS '日期',
- new.cnt AS '当日注册数',
- out.cnt AS '次日留存数',
- CAST(out.cnt*100.0/new.cnt AS varchar(20))+'%' AS '留存率'
- FROM (SELECT COUNT(1) as cnt, CONVERT(nvarchar(20),register_time,111) AS register_day
- FROM Retention
- GROUP BY CONVERT(nvarchar(20),register_time,111)
- ) AS new
- JOIN
- (SELECT COUNT(1) as cnt, CONVERT(nvarchar(20),register_time,111) AS register_day
- FROM Retention
- where CONVERT(nvarchar(20),logout_time,111) > CONVERT(nvarchar(20),DATEADD(DAY,1,register_time),111)
- GROUP BY CONVERT(nvarchar(20),register_time,111)
- ) AS out
- ON new.register_day=out.register_day
|
计算结果:

这个方法好麻烦,有没有简单一点的方法/(ㄒoㄒ)/~~
5、删除表