小编典典

在六个月内保留的参与者百分比

sql

我是一位学校老师，对MS SQL Server非常陌生。每个人都建议尝试此站点。开始！

我正在尝试编写查询，以测试针对学术课程参与程度的不同类型的结果度量。我想尝试几种计算此结果度量的方法。我要计算的结果是：
在该计划的六个月中保留的参与者百分比是多少？ 我正在测试定义参与者和不同时间范围的不同方法。我正在尝试产生4个查询。不幸的是，我必须使用不同的表：
出勤率，状态，取消注册，无效 。我从下面列出了每个示例数据

查询

参与者被定义为大家，至少参加一类，每周两次为6个月（第181天累积的总）开始于2012年7月1日和结束2013年6月30日，所以财政year.If长度的参与者 deenrolled 或 不活动 ，他们被丢弃。
参与者定义为从2013年1月1日开始，每周至少参加两次课程，共6个月（共181天）的 所有人 。如果参与者 取消注册 或变得 不活跃， 他们将被放弃。
从2013年1月1日到今天，每周至少参加两次课程的每个人都定义为参加者
参与者被定义为学生的入学开始日期，直到他们被取消注册或变得不活跃为止。

参与者（分子）参与者/所服务的所有学生（分母）

我要查找的4个查询输出是此版本的不同版本：

例子

Participants    Served   Percent_Served
75               100        75%

我一直在下面弄乱查询的不同版本

 SELECT 
Count (distinct ID) as Count, 
  Count  ( DATEADD( dd, -181, DATEADD(wk, DATEDIFF(wk,0,Date), 0)) > 2 as Participants ,
FROM Attendance
where Attendence_date date between '07/01/2012' and '06/30/2013'
and ID not in (Select ID from Inactive) 
or ID not in (select ID from Deenrolled) 
GROUP BY ID

和

 SELECT 
Count (distinct ID) as Count, 
  Count  ( DATEADD( dd, -181, DATEADD(wk, DATEDIFF(wk,0,Date), 0)) - Enrolled_Date  as Participants ,
FROM Attendance
where Attendence_date date between '07/01/2012' and '06/30/2013'
and ID not in (Select ID from Inactive) 
or ID not in (select ID from Deenrolled) 
GROUP BY ID

对于这些查询的任何编程帮助将不胜感激。

以下是样本/示例数据集。

Attendence_date是学生参加一个班级的日期。

CREATE TABLE Attendance (
    ID int,
    Attendence_date datetime
    )

INSERT INTO Attendance VALUES 
(4504498,  '7/1/2012'),
(4504498,  '7/2/2012'),
(4504498,   '7/3/2012'),
(4504498,   '7/4/2012'),
(4504498,   '7/5/2012'),
(4504498,   '7/8/2012'),
(4504498,   '7/9/2012'),
(4504498,   '7/10/2012'),
(4504498,   '7/11/2012'),
(4504498,   '7/12/2012'),
(4504498,   '7/1/2012'),
(4504498,   '7/2/2012'),
(4504498,   '7/3/2012'),
(4504498,   '7/4/2012'),
(4504498,   '7/5/2012'),
(4504498,   '7/8/2012'),
(4504498,   '7/9/2012'),
(4504498,   '7/10/2012'),
(4504498,   '7/11/2012'),
(4504498,   '7/12/2012'),
(9201052,   '7/15/2012'),
(9201052,   '7/16/2012'),
(9201052,   '7/17/2012'),
(9201052,   '7/17/2012'),
(9201052,   '7/18/2012'),   
(7949745,   '7/17/2012'),   
(7949745,   '7/18/2012'),
(7949745,   '7/23/2012'),   
(7949745,   '7/23/2012'),   
(7949745,   '7/24/2012'),
(7949745,   '7/26/2012'),
(7949745,   '7/26/2012'),   
(7949745,   '8/8/2012'),    
(7949745,   '8/8/2012'),    
(7949745,   '11/5/2012'),   
(7949745,   '11/5/2012'),   
(7949745,   '11/5/2012'),   
(7949745,   '11/6/2012'),   
(7949745,   '11/6/2012'),   
(7949745,   '11/6/2012'),   
(7949745,   '11/7/2012'),   
(7949745,   '11/7/2012'),   
(7949745,   '11/7/2012')

这是包含注册日期的信息。

CREATE TABLE [Status] (
    ID int,
    Intake_Date datetime ,
   Engaged_Date datetime ,
   Enrolled_Date datetime)
INSERT INTO [Status] VALUES 
(7949745, '3/7/2012',   '7/17/2012', '3/8/2012'),
(4504498, '2/21/2013',  '3/5/2013',  '3/22/2013'),
(1486279, '4/18/2013',  '5/7/2013',   '5/20/2013'),
(9201052, '5/15/2012',  '7/13/2012',  '5/15/2012'),
(1722390, '3/5/2012',   '8/27/2012', '3/8/2012'),
(7735695, '9/7/2012',   '9/7/2012',  '9/28/2012'),
(9261549, '3/7/2012',   '7/24/2012', '3/8/2012'),
(3857008, '3/15/2013',  '3/18/2013', '4/3/2013'),
(8502583, '3/14/2013',     '4/15/2013', '5/3/2013'),
(1209774,  '4/19/2012',  '1/1/2012',   '4/24/2012')

这是包含取消注册日期的信息。

CREATE TABLE Deenrolled (
    ID int,
    Deenrolled_Date datetime)
INSERT INTO Deenrolled  VALUES 
(7949745,    '2/4/2013'),
(5485272,    '07/08/2013'),
(8955628,    '01/10/2013'),
(5123221,    '7/8/2013'),
(5774753,    '7/18/2013'),
(3005451,    '2/18/2013'),
(7518818,    '05/29/2013'),
(9656985,    '6/20/2013'),
(2438101,    '7/17/2013'),
(1437052,    '7/25/2013'),
(9133874,    '4/25/2013'),
(7007375,    '6/19/2013'),
(3178181,    '5/24/2013')

而且不活跃

CREATE TABLE Inactive (
    ID int,
   Effect_Date datetime)
INSERT INTO Inactive VALUES 
(1209774,       '10/12/2012'),
(5419494,       '10/12/2012'),
(4853049,       '10/9/2012'),
(1453678,       '5/23/2013'),
(1111554,       '7/16/2012'),
(5564128,       '2/15/2013'),
(1769234,       '7/16/2012')

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2021-04-22

共1个答案

小编典典

好吧，这不是一件容易的事。主要问题是要解决“六个月至少每周两次”的部分-每周计算两次很容易，但是应该连续6个月！

在尝试解决该问题时，我找到了尼尔斯·范德·雷斯特（Niels van derRest）的绝妙答案-在一组数字中找到连续范围。因此，我将为您提供
第1部分的 一般查询，您可以更改参数并获取 第2部分的 结果：

declare @Weeks int, @PerWeek int, @StartDate date, @EndDate date, @count

select
    @StartDate = '20120701',
    @EndDate = '20130630',
    @Weeks = 26, -- 6 month or 26 weeks
    @PerWeek = 2 -- twice per week

select @count = count(distinct A.ID)
from Attendance as A
where
    A.Attendence_date between @StartDate and @EndDate and
    A.ID not in (select T.ID from Deenrolled as T) and
    A.ID not in (select T.ID from Inactive as T)

;with CTE as (
    -- Week numbers, filter by dates
    select
        A.ID,
        datediff(dd, @StartDate, A.Attendence_date) / 7 as Wk
    from Attendance as A
    where
        A.Attendence_date between @StartDate and @EndDate and
        A.ID not in (select T.ID from Deenrolled as T) and
        A.ID not in (select T.ID from Inactive as T)
  ), CTE2 as (
    -- Group by week, filter less then @PerWeek per week, calculate row number
    select
        Wk, ID,
        row_number() over (partition by ID order by Wk) as Row_Num
    from CTE
    group by Wk, ID
    having count(*) >= @PerWeek
)
-- Final query - group by difference between week and row_number
select 100 * cast(count(distinct ID) as float) / @count
from CTE2
group by ID, Wk - Row_Num
having count(*) >= @Weeks

我创建了 SQL FIDDLE EXAMPLE ，您可以测试查询。

第3部分 很简单

declare @PerWeek int, @StartDate date

select
    @StartDate = '20130101',
    @PerWeek = 2 -- twice per week

select @count = count(distinct A.ID)
from Attendance as A
where
    A.Attendence_date >= @StartDate and
    A.ID not in (select T.ID from Deenrolled as T) and
    A.ID not in (select T.ID from Inactive as T)

;with CTE as (
    -- Week numbers, filter by dates
    select
        A.ID,
        datediff(dd, @StartDate, A.Attendence_date) / 7 as Wk
    from Attendance as A
    where
        A.Attendence_date >= @StartDate and
        A.ID not in (select T.ID from Deenrolled as T) and
        A.ID not in (select T.ID from Inactive as T)
  ), CTE2 as (
    -- Group by week, filter less then @PerWeek per week
    select distinct ID
    from CTE
    group by Wk, ID
    having count(*) >= @PerWeek
)
select 100 * cast(count(*) as float) / @count from CTE2

第4部分 对我来说似乎有点不清楚，您能澄清一下吗？

2021-04-22