1. 數據準備
-- 數據準備
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: 差值計算
-- 1. 對用戶數據進行分組,按照活躍日期進行排序(去重:防止有一天有多次活躍記錄)
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_id | active_date | rn |
|---|
| 10001 | 2023-02-01 | 1 |
| 10001 | 2023-02-03 | 2 |
| 10001 | 2023-02-04 | 3 |
| 10001 | 2023-02-05 | 4 |
| 10002 | 2023-02-02 | 1 |
| 10002 | 2023-02-03 | 2 |
| 10002 | 2023-02-04 | 3 |
| 10002 | 2023-02-05 | 4 |
| 10002 | 2023-02-07 | 5 |
| … | … | … |
-- 2. 使用活躍日期和排序rn進行差值計算,得到的日期如果是相等的,就說明活躍日期是連續(xù)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_id | active_date | rn | sub_date |
|---|
| 10001 | 2023-02-01 | 1 | 2023-01-31 |
| 10001 | 2023-02-03 | 2 | 2023-02-01 |
| 10001 | 2023-02-04 | 3 | 2023-02-01 |
| 10001 | 2023-02-05 | 4 | 2023-02-01 |
| 10002 | 2023-02-02 | 1 | 2023-02-01 |
| 10002 | 2023-02-03 | 2 | 2023-02-01 |
| 10002 | 2023-02-04 | 3 | 2023-02-01 |
| 10002 | 2023-02-05 | 4 | 2023-02-01 |
| 10002 | 2023-02-07 | 5 | 2023-02-02 |
| … | … | … | … |
-- 3. 按照user_id和sub_date 進行分組求和,篩選出連續(xù)登陸天數大于3天的用戶
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_id | begin_date | end_date | login_duration |
|---|
| 10001 | 2023-02-03 | 2023-02-05 | 3 |
| 10002 | 2023-02-02 | 2023-02-05 | 4 |
| 10003 | 2023-02-02 | 2023-02-08 | 7 |
| 10004 | 2023-02-06 | 2023-02-08 | 3 |
3. 方法二: lead或lag函數
-- 1. 將active_date 上抬2行,不存在默認為'0'(計算連續(xù)活躍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躍記錄)
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_id | active_date | lead_active_date |
|---|
| 10001 | 2023-02-01 | 2023-02-04 |
| 10001 | 2023-02-03 | 2023-02-05 |
| 10001 | 2023-02-04 | 0 |
| 10001 | 2023-02-05 | 0 |
| 10002 | 2023-02-02 | 2023-02-04 |
| 10002 | 2023-02-03 | 2023-02-05 |
| 10002 | 2023-02-04 | 2023-02-07 |
| 10002 | 2023-02-05 | 0 |
| 10002 | 2023-02-07 | 0 |
| … | … | … |
-- 2. 過濾篩選出, lead_active_date 與 active_date 差值為2的, 差值2 -> 連續(xù)活躍了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_id | active_date | lead_active_date |
|---|
| 10001 | 2023-02-03 | 2023-02-05 |
| 10002 | 2023-02-02 | 2023-02-04 |
| 10002 | 2023-02-03 | 2023-02-05 |
| … | … | … |
-- 3. user_id 去重, 得到連續(xù)活躍天數>=3天的用戶
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_id |
|---|
| 10001 |
| 10002 |
| 10003 |
| 10004 |
到此這篇關于SQL統(tǒng)計連續(xù)登陸3天用戶的實現示例的文章就介紹到這了,更多相關SQL統(tǒng)計連續(xù)登陸3天用戶內容請搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關文章希望大家以后多多支持腳本之家!