如何在同一张表上没有左连接的情况下获得 D1 D7 D30

Question

目标：

我想知道在第 0 天注册的每个人，有多少人在 D1、D7 和 D30 之后登录。我想做一个表值函数，用户在其中插入日期并获取 D0、D1、D7、D30 的结果。 结果应如下所示：

Date        TotalD0 TotalD1 TotalD7 TotalD30
2019-04-01    3         3      2       1

情况：

我有一个包含电子邮件和 login_time 的登录表。我在同一张桌子上离开了三次，它在一张假桌子上工作。但是，当使用具有数百万行的真实数据时，它会永远运行。必须有一种更有效的方法来做到这一点。

我尝试了什么：

CREATE FUNCTION fnTestData
(
    @StartDate AS Date
)
RETURNS TABLE
AS
RETURN
    select @startdate, 
       COUNT(distinct t1.id) As TotalD0, 
       COUNT(distinct t1a.id) As TotalD1,
       COUNT(distinct t1b.id) As TotalD7,
       COUNT(distinct t1c.id) As TotalD30
    from #test1 t1
       left join #test1 t1a on t1.id=t1a.id and t1a.login_time >= 
             DATEADD(day,1,t1.login_time)
       left join #test1 t1b on t1.id=t1b.id and t1b.login_time >= 
             DATEADD(day,7,t1.login_time)
       left join #test1 t1c on t1.id=t1c.id and t1c.login_time >= 
             DATEADD(day,30,t1.login_time)
    where t1.login_time = @startdate
    group by t1.login_time

测试数据：

create table #test1 (id int, login_time date)
insert into #test1 values
(1, '2019-04-01'),
(1, '2019-04-01'),
(1, '2019-04-02'),
(1, '2019-04-19'),
(1, '2019-05-05'),
(2, '2019-04-01'),
(2, '2019-04-05'),
(2, '2019-04-10'),
(2, '2019-04-15'),
(3, '2019-04-01'),
(3, '2019-04-01'),
(3, '2019-04-02')

Answer 1

您的查询可以翻译成GROUP BY：

DECLARE @StartDate Date = '2019-04-01'

SELECT COUNT(DISTINCT id) D0
     , COUNT(DISTINCT CASE WHEN login_time >= DATEADD(DAY,  1, @StartDate) THEN id END) AS D1
     , COUNT(DISTINCT CASE WHEN login_time >= DATEADD(DAY,  7, @StartDate) THEN id END) AS D7
     , COUNT(DISTINCT CASE WHEN login_time >= DATEADD(DAY, 30, @StartDate) THEN id END) AS D30
FROM #test1 AS t
WHERE login_time >= @StartDate
AND EXISTS (
    SELECT 1
    FROM #test1 AS x
    WHERE x.id = t.id
    AND x.login_time = @StartDate
)

D0    D1    D7    D30
3     3     2     1

您需要创建适当的索引来加快速度。

Answer 2

如果您想根据人们开始的日期进行同期群分析：

select first_ld,
       count(*) as num_d0,
       sum(case when login_date >= dateadd(day, 1, firstld) then 1 else 0 end) as num_d1,
       sum(case when login_date >= dateadd(day, 7, firstld) then 1 else 0 end) as num_d7,
       sum(case when login_date >= dateadd(day, 30, firstld) then 1 else 0 end) as num_d30
from (select id, convert(date, login_time) as login_date,
             min(convert(date, login_time)) over (partition by id) as first_ld
      from #test1 t
      group by id, convert(date, login_time)
     ) t
group by first_ld
order by first_ld;