用户留存率是什么?如何用sql求出次日留存率?

本文通过SQL语句展示了如何创建临时表并提取用户登录数据,进而计算次日留存用户、留存率,并详细解析了每日次日留存率的计算过程。关键步骤包括筛选次日登录用户、使用DATEDIFF函数比较日期,以及计算百分比。

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建表sql语句:

drop table if exists liucunlv;
create table liucunlv as select * from ( 
select 'a' as user_name, '2021-03-01'as logindate
 union all 
select 'a' as user_name, '2021-03-02'as logindate
 union all 
select 'a' as user_name, '2021-03-03'as logindate
 union all 
select 'a' as user_name, '2021-03-05'as logindate
 union all 
select 'b' as user_name, '2021-03-02'as logindate
 union all 
select 'b' as user_name, '2021-03-03'as logindate
 union all 
select 'c' as user_name, '2021-03-02'as logindate
 union all 
select 'c' as user_name, '2021-03-04'as logindate
 union all 
select 'c' as user_name, '2021-03-05'as logindate
 union all 
select 'c' as user_name, '2021-03-06'as logindate
 union all 
select 'c' as user_name, '2021-03-07'as logindate
 union all 
select 'd' as user_name, '2021-03-07'as logindate) t
SELECT * FROM liucunlv

在这里插入图片描述
一:现在要求出哪些用户才是次日留存的用户

SELECT user_name,MIN(logindate) AS newdate FROM liucunlv GROUP BY user_name

在这里插入图片描述
如果logindate和newdate相差一天,那么就让if_ciriliucun=1否则if_ciriliucun=0

SELECT t1.user_name,CASE WHEN DATEDIFF(logindate,newdate)=1 THEN 1 ELSE 0 END AS if_ciriliucun
FROM 
(
SELECT user_name,MIN(logindate) AS newdate FROM liucunlv GROUP BY user_name
)t1
JOIN liucunlv t2 ON t1.user_name=t2.user_name

在这里插入图片描述
之后再做一个groupby user_name,就能选出是否是次日留存的用户了

SELECT t3.user_name,MAX(if_ciriliucun) FROM
(
SELECT t1.user_name,CASE WHEN DATEDIFF(logindate,newdate)=1 THEN 1 ELSE 0 END AS if_ciriliucun
FROM 
(
SELECT user_name,MIN(logindate) AS newdate FROM liucunlv GROUP BY user_name
)t1
JOIN liucunlv t2 ON t1.user_name=t2.user_name)t3 GROUP BY t3.user_name

在这里插入图片描述
可以看到,a和b是次日留存的用户

二:求每日的次日留存率
留存率解释:
在互联网行业中,用户在某段时间内开始使用应用,经过一段时间后,仍然继续使用该应用的用户,被认作是留存用户。这部分用户占当时新增用户的比例即是留存率,会按照每隔1单位时间(例日、周、月)来进行统计。

SELECT t1.logindate AS 日期,
       COUNT(DISTINCT CASE WHEN DATEDIFF(t3.logindate,newdate) = 1 THEN t3.user_name ELSE NULL END) AS 次日留存用户数,
       COUNT(DISTINCT t2.user_name) AS 新增用户数,
       CONCAT(round(COUNT(DISTINCT CASE WHEN DATEDIFF(t3.logindate,newdate) = 1 THEN t3.user_name ELSE NULL END)*100/COUNT(DISTINCT t2.user_name),2),'%') AS 次日留存率
FROM
  (SELECT DISTINCT logindate
   FROM liucunlv) t1
LEFT JOIN
  (SELECT user_name,
          MIN(logindate) newdate
   FROM liucunlv
   GROUP BY user_name) t2 ON t2.newdate = t1.logindate
LEFT JOIN liucunlv t3 ON t2.user_name = t3.user_name
GROUP BY t1.logindate

在这里插入图片描述

求每日的次日留存率的整体代码如果读起来有困难,可以尝试接着往下读:
将求每日的次日留存率的整体代码拆解如下::
t1:

SELECT DISTINCT logindate
   FROM liucunlv

在这里插入图片描述
t2:

SELECT user_name,
          MIN(logindate) newdate
   FROM liucunlv
   GROUP BY user_name

在这里插入图片描述
t3:

在这里插入图片描述
将t1,t2,t3合起来得到一个表a

SELECT *
FROM
  (SELECT DISTINCT logindate
   FROM liucunlv) t1
LEFT JOIN
  (SELECT user_name,
          MIN(logindate) newdate
   FROM liucunlv
   GROUP BY user_name) t2 ON t2.newdate = t1.logindate
LEFT JOIN liucunlv t3 ON t2.user_name = t3.user_name

在这里插入图片描述
再将表a group by t1.logindate (表a的第一列),做一些聚合运算,便得到

SELECT t1.logindate AS 日期,
       COUNT(DISTINCT CASE WHEN DATEDIFF(t3.logindate,newdate) = 1 THEN t3.user_name ELSE NULL END) AS 次日留存用户数,
       COUNT(DISTINCT t2.user_name) AS 新增用户数,
       CONCAT(round(COUNT(DISTINCT CASE WHEN DATEDIFF(t3.logindate,newdate) = 1 THEN t3.user_name ELSE NULL END)*100/COUNT(DISTINCT t2.user_name),2),'%') AS 次日留存率
FROM
  (SELECT DISTINCT logindate
   FROM liucunlv) t1
LEFT JOIN
  (SELECT user_name,
          MIN(logindate) newdate
   FROM liucunlv
   GROUP BY user_name) t2 ON t2.newdate = t1.logindate
LEFT JOIN liucunlv t3 ON t2.user_name = t3.user_name
GROUP BY t1.logindate

得到我们想要的次日留存率:
在这里插入图片描述

引用:
https://blog.youkuaiyun.com/weixin_38617657/article/details/116697972

### SQL 次日留存率计算方法及面试题解答 次日留存率是衡量用户在首次登录后的第二天是否继续使用产品的指标。它对于评估产品的吸引力和用户的粘性具有重要意义[^2]。以下是次日留存率的计算方法及相关SQL面试题的解答。 #### 1. 次日留存率的定义 次日留存率是指在某一天新增的用户中,有多少比例的用户在第二天再次登录系统。公式如下: \[ 次日留存率 = \frac{次日再次登录的用户数}{新增用户数} \times 100\% \] #### 2. 计算次日留存率SQL实现 以下是一个通用的SQL实现方式,用于计算次日留存率: ```sql WITH FirstLogin AS ( -- 找出每个用户的首次登录时间 SELECT user_id, MIN(record_time) AS first_record_time FROM user_log GROUP BY user_id ), RetentionUsers AS ( -- 找出次日登录的用户 SELECT a.user_id, a.record_time FROM user_log a JOIN FirstLogin b ON a.user_id = b.user_id WHERE DATE(a.record_time) = DATE(DATE_ADD(b.first_record_time, INTERVAL 1 DAY)) ) -- 计算次日留存率 SELECT COUNT(DISTINCT RetentionUsers.user_id) AS next_day_retention_users, COUNT(DISTINCT FirstLogin.user_id) AS initial_users, ROUND(COUNT(DISTINCT RetentionUsers.user_id) / COUNT(DISTINCT FirstLogin.user_id) * 100, 2) AS next_day_retention_rate FROM FirstLogin LEFT JOIN RetentionUsers ON FirstLogin.user_id = RetentionUsers.user_id; ``` 上述代码通过两个CTE(Common Table Expressions)分别找出首次登录的用户次日再次登录的用户,然后通过连接操作计算次日留存率[^4]。 #### 3. 面试题示例 **题目:** 给定一张记录用户登录信息的表`user_log`,包含字段`user_id`(用户ID)和`record_time`(登录时间)。请编写SQL查询,计算每天新增用户次日留存率。 **解答:** ```sql WITH DailyNewUsers AS ( -- 找出每天新增的用户 SELECT user_id, record_time AS first_login_date FROM ( SELECT user_id, record_time, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY record_time) AS rn FROM user_log ) t WHERE rn = 1 ), NextDayLogins AS ( -- 找出次日登录的用户 SELECT a.user_id, a.record_time FROM user_log a JOIN DailyNewUsers b ON a.user_id = b.user_id WHERE DATE(a.record_time) = DATE(DATE_ADD(b.first_login_date, INTERVAL 1 DAY)) ) -- 计算每天新增用户次日留存率 SELECT DATE(b.first_login_date) AS login_date, COUNT(DISTINCT a.user_id) AS next_day_retention_users, COUNT(DISTINCT b.user_id) AS new_users, ROUND(COUNT(DISTINCT a.user_id) / COUNT(DISTINCT b.user_id) * 100, 2) AS next_day_retention_rate FROM NextDayLogins a RIGHT JOIN DailyNewUsers b ON a.user_id = b.user_id GROUP BY DATE(b.first_login_date); ``` 此查询首先通过窗口函数`ROW_NUMBER()`找出每天新增的用户,然后计算这些用户次日再次登录的比例[^3]。 --- ### 知识点总结 - **时间函数:** 使用`MIN()`、`DATE()`、`DATE_ADD()`等函数处理日期数据。 - **窗口函数:** 使用`ROW_NUMBER()`标记每个用户的首次登录记录。 - **连接操作:** 使用`JOIN`或`LEFT JOIN`将首次登录用户次日登录用户进行匹配。 - **聚合函数:** 使用`COUNT()`统计用户数量,并结合`ROUND()`函数计算百分比。 ---
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