SQL统计连续登陆3天的用户(连续活跃超3天用户)
目录
SQL统计连续登陆3天的用户(连续活跃超3天用户)1. 数据准备2. 方法一: 差值计算3. 方法二: lead或lag函数end
1. 数据准备
-- 数据准备
WITH user_active_info AS (
SELECT * FROM (
VALUES ('10001' , '2023-02-01'),('10001' , '2023-02-03')
,('10001' , '2023-02-04'),('10001' , '2023-02-05')
,('10002' , '2023-02-02'),('10002' , '2023-02-03')
,('10002' , '2023-02-04'),('10002' , '2023-02-05')
,('10002' , '2023-02-07'),('10003' , '2023-02-02')
,('10003' , '2023-02-03'),('10003' , '2023-02-04')
,('10003' , '2023-02-05'),('10003' , '2023-02-06')
,('10003' , '2023-02-07'),('10003' , '2023-02-08')
,('10004' , '2023-02-03'),('10004' , '2023-02-04')
,('10004' , '2023-02-06'),('10004' , '2023-02-07')
,('10004' , '2023-02-08'),('10004' , '2023-02-08')
,('10005' , '2023-02-02'),('10005' , '2023-02-05')
) AS user_active_info(user_id, active_date)
)
2. 方法一: 差值计算
-- 1. 对用户数据进行分组,按照活跃日期进行排序(去重:防止有一天有多次活跃记录)
SELECT
user_id
, active_date
, ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY active_date) AS rn
FROM user_active_info
GROUP BY user_id , active_date
;
user_idactive_datern100012023-02-011100012023-02-032100012023-02-043100012023-02-054100022023-02-021100022023-02-032100022023-02-043100022023-02-054100022023-02-075………
-- 2. 使用活跃日期和排序rn进行差值计算,得到的日期如果是相等的,就说明活跃日期是连续的
SELECT
user_id
, active_date
, rn
, DATE_SUB(active_date,rn) AS sub_date
FROM (
SELECT
user_id
, active_date
, ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY active_date) AS rn
FROM user_active_info
GROUP BY user_id , active_date
) a
;
user_idactive_daternsub_date100012023-02-0112023-01-31100012023-02-0322023-02-01100012023-02-0432023-02-01100012023-02-0542023-02-01100022023-02-0212023-02-01100022023-02-0322023-02-01100022023-02-0432023-02-01100022023-02-0542023-02-01100022023-02-0752023-02-02…………
-- 3. 按照user_id和sub_date 进行分组求和,筛选出连续登陆天数大于3天的用户
SELECT
user_id
, MIN(active_date) AS begin_date
, MAX(active_date) AS end_date
, COUNT (1) AS login_duration
FROM (
SELECT
user_id
, active_date
, rn
, DATE_SUB(active_date,rn) AS sub_date
FROM (
SELECT
user_id
, active_date
, ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY active_date) AS rn
FROM user_active_info
GROUP BY user_id , active_date
) a
) b
GROUP BY user_id , sub_date
HAVING login_duration >= 3
;
user_idbegin_dateend_datelogin_duration100012023-02-032023-02-053100022023-02-022023-02-054100032023-02-022023-02-087100042023-02-062023-02-083
3. 方法二: lead或lag函数
-- 1. 将active_date 上抬2行,不存在默认为'0'(计算连续活跃3天以上的, 上抬2行,n天上抬n-1行)(去重:防止有一天有多次活跃记录)
SELECT
user_id
, active_date
, lead(active_date , 2 , 0) OVER(PARTITION BY user_id ORDER BY active_date) AS lead_active_date
FROM user_active_info
GROUP BY user_id , active_date
user_idactive_datelead_active_date100012023-02-012023-02-04100012023-02-032023-02-05100012023-02-040100012023-02-050100022023-02-022023-02-04100022023-02-032023-02-05100022023-02-042023-02-07100022023-02-050100022023-02-070………
-- 2. 过滤筛选出, lead_active_date 与 active_date 差值为2的, 差值2 -> 连续活跃了3天
SELECT
user_id , active_date , lead_active_date
FROM (
SELECT
user_id
, active_date
, lead(active_date , 2 , 0) OVER(PARTITION BY user_id ORDER BY active_date) AS lead_active_date
FROM user_active_info
GROUP BY user_id , active_date
) a
WHERE lead_active_date != '0'
AND DATEDIFF(lead_active_date , active_date) = 2
user_idactive_datelead_active_date100012023-02-032023-02-05100022023-02-022023-02-04100022023-02-032023-02-05………
-- 3. user_id 去重, 得到连续活跃天数>=3天的用户
SELECT
user_id
FROM (
SELECT
user_id , active_date , lead_active_date
FROM (
SELECT
user_id
, active_date
, lead(active_date , 2 , 0) OVER(PARTITION BY user_id ORDER BY active_date) AS lead_active_date
FROM user_active_info
GROUP BY user_id , active_date
) a
WHERE lead_active_date != '0'
AND DATEDIFF(lead_active_date , active_date) = 2
) b
GROUP BY user_id
user_id10001100021000310004
end
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