【sql】查询每日活跃用户数/每日新增用户数/每日累计用户数

建表

CREATE TABLE denglu(
id varchar(10),
dt date)

插入测试数据

INSERT INTO denglu 
VALUES 
('01','2021-10-01'),
('01','2021-10-01'),
('02','2021-10-03'),
('02','2021-10-02'),
('03','2021-10-01'),
('03','2021-10-04'),
('04','2021-10-06'),
('05','2021-10-07'),
('06','2021-10-06'),
('06','2021-10-05'),
('06','2021-10-04');

Q1 每日活跃用户数
按天数统计，需要group by dt，一个用户一天内可能多次登陆，需要distinct去重，语法如下

SELECT dt,count(DISTINCT id) FROM denglu 
GROUP BY dt ORDER BY dt

Q2 每日新增用户数
该日新增用户，即在此日期之前没有出现过的用户，因此每个用户的最小登陆日期计为那一日的新增，语法如下

SELECT dt2,count(DISTINCT id) FROM (
SELECT id,min(dt)AS dt2 FROM denglu GROUP BY id
)
GROUP BY dt2

Q3 每日累计用户数
先放语法

SELECT dt2,count(DISTINCT CASE WHEN dt1<=dt2 THEN id end) FROM (
(SELECT distinct dt AS dt1,id FROM denglu) a
LEFT JOIN (SELECT DISTINCT dt AS dt2 FROM denglu)b ON 1=1
)
GROUP BY dt2

拆解，先找出每日登陆用户

SELECT distinct dt AS dt1,id FROM denglu

DT1	ID
2021-10-01	03
2021-10-01	01
2021-10-02	02
2021-10-03	02
2021-10-04	03
2021-10-04	06
2021-10-05	06
2021-10-06	04
2021-10-06	06
2021-10-07	05

第二步，左关联一个distinct去重的日期，方便后面统计每日累计

SELECT * FROM (
(SELECT distinct dt AS dt1,id FROM denglu) a
LEFT JOIN (SELECT DISTINCT dt AS dt2 FROM denglu)b ON 1=1
)ORDER BY DT1
FETCH FIRST 20 ROWS ONLY;

DT1	ID	DT2
2021-10-01	01	2021-10-01
2021-10-01	01	2021-10-02
2021-10-01	01	2021-10-03
2021-10-01	01	2021-10-04
2021-10-01	01	2021-10-05
2021-10-01	01	2021-10-06
2021-10-01	01	2021-10-07
2021-10-01	03	2021-10-01
2021-10-01	03	2021-10-02
2021-10-01	03	2021-10-03
2021-10-01	03	2021-10-04
2021-10-01	03	2021-10-05
2021-10-01	03	2021-10-06
2021-10-01	03	2021-10-07
2021-10-02	02	2021-10-01
2021-10-02	02	2021-10-02
2021-10-02	02	2021-10-03
2021-10-02	02	2021-10-04
2021-10-02	02	2021-10-05
2021-10-02	02	2021-10-06

之后就以DT2为每日统计，按天数统计，group by DT2，当DT1<=DT2，则计为DT2累计用户