确定客户的访问次数和停留时间

Question

我有一组只有 ClientID 和 ServiceDate 的客户记录

我想确定每次访问的访问次数和停留时间

我已经在 MS Access 中创建了一些代码并阅读了搜索此问题时推荐的几个类似问题以及从 Google 搜索

返回的其他建议

当前代码：

SELECT 
  a.Client, 
  a.ServiceDate, 
  DateAdd ( "d", -1, a.ServiceDate ) AS [Prior], 
  DateAdd ( "d", 1, a.ServiceDate ) AS [Next], ( SELECT COUNT(*)
FROM Services_Table b
WHERE a.Client = b.Client 
      AND DateAdd ( "d", -1, a.ServiceDate ) = b.ServiceDate ) AS [Index]
FROM Services_Table AS a
ORDER BY a.Client, a.ServiceDate;

输入数据：

Client  ServiceDate
1   8/4/2018
1   8/5/2018
1   8/6/2018
6   7/11/2018
6   7/12/2018
6   7/17/2018
6   7/18/2018
6   7/19/2018
6   7/30/2018
6   7/31/2018
6   8/1/2018
6   8/2/2018
15  1/13/2019
15  1/14/2019
5   9/13/2018
5   9/14/2018
5   9/15/2018

当前输出：

Clt ServiceDate Prior           Next          Index
1   8/4/2018    8/3/2018    8/5/2018    0
1   8/5/2018    8/4/2018    8/6/2018    1
1   8/6/2018    8/5/2018    8/7/2018    1
5   9/13/2018   9/12/2018   9/14/2018   0
5   9/14/2018   9/13/2018   9/15/2018   1
5   9/15/2018   9/14/2018   9/16/2018   1
6   7/11/2018   7/10/2018   7/12/2018   0
6   7/12/2018   7/11/2018   7/13/2018   1
6   7/17/2018   7/16/2018   7/18/2018   0
6   7/18/2018   7/17/2018   7/19/2018   1
6   7/19/2018   7/18/2018   7/20/2018   1
6   7/30/2018   7/29/2018   7/31/2018   0
6   7/31/2018   7/30/2018   8/1/2018    1
6   8/1/2018    7/31/2018   8/2/2018    1
6   8/2/2018    8/1/2018    8/3/2018    1
15  1/13/2019   1/12/2019   1/14/2019   0
15  1/14/2019   1/13/2019   1/15/2019   1

期望的结果：

Clt Start           End     Length
1   8/4/2018    8/6/2018      3
5   9/13/2018   9/15/2018     3
6   7/11/2018   7/12/2018     2
6   7/17/2018   7/19/2018     3
6   7/30/2018   8/2/2018      4
15  1/13/2019   1/14/2019     2

Answer 1

让我们从确定起点开始：

开始日期是没有连续的前一个日期的日期，因此：

SELECT Client, ServiceDate
FROM Services_Table t
WHERE NOT EXISTS(SELECT 1 FROM Services_Table s WHERE s.Client = t.Client AND s.ServiceDate = DateAdd("d", -1, t.ServiceDate))

同样，我们可以识别两端：

SELECT Client, ServiceDate
FROM Services_Table t
WHERE NOT EXISTS(SELECT 1 FROM Services_Table s WHERE s.Client = t.Client AND s.ServiceDate = DateAdd("d", 1, t.ServiceDate))

然后，我们可以将开始和结束连接在一起，select最接近开始的结束日期：

SELECT Start.Client, Start.ServiceDate As Start, Min(End.ServiceDate) As End, Min(End.ServiceDate) - Start.ServiceDate As Length
FROM
(
    SELECT Client, ServiceDate
    FROM Services_Table t
    WHERE NOT EXISTS(SELECT 1 FROM Services_Table s WHERE s.Client = t.Client AND s.ServiceDate = DateAdd("d", -1, t.ServiceDate))
) Start
INNER JOIN (
    SELECT Client, ServiceDate
    FROM Services_Table t
    WHERE NOT EXISTS(SELECT 1 FROM Services_Table s WHERE s.Client = t.Client AND s.ServiceDate = DateAdd("d", 1, t.ServiceDate))
) End ON (Start.Client = End.Client AND Start.ServiceDate <= End.ServiceDate)
GROUP BY Start.Client, Start.ServiceDate

确定客户的访问次数和停留时间

Determine Number of Visits and Length of Stay by Client

sql-server

ms-access

sql-server-2016

ms-access-2016