连续的日期 [英] Consecutive streak of dates

问题描述

希望这不是另一个问题的欺骗，但我在其他任何地方都看不到它 - 这也是 我问的另一个问题，希望能让我开始研究如何处理它.

Hopefully this isn't a dupe of another question, but I couldn't see it anywhere else - also this is a simplified version of another question I asked, hopefully to get me started on working out how to approach it.

我希望计算出每个月至少有一次付款的连续付款范围.

I am looking to work out consecutive ranges of payments where there has been at least one payment in each month.

我有以下示例数据

CREATE TABLE #data
(
Contact_reference NVARCHAR(55)
,Date_payment DATETIME
,Payment_value MONEY
)

INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2003-06-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2004-06-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2004-12-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-04-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-05-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-06-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-07-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-08-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-09-08',12.82)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-10-10',12.8205)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-11-10',12.8205)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-12-10',12.8205)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-01-10',12.8205)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-02-10',12.8205)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-02-28',12.8205)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-04-12',12.8205)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-05-10',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-06-11',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-07-10',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-08-09',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-09-10',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-10-09',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-11-09',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-12-10',19.2308)
INSERT INTO #data VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2008-01-10',19.2308)

我希望能够为每个联系人计算出他们连续提供的范围(定义为每个日历月至少提供一次)、连续付款的次数、每次付款的总价值范围(如果可能的话，最好是当前范围与最近一个范围结束之间的差距).

And what I would like to be able to do is to work out for each contact the ranges over which they gave consecutively (defined as giving at least once in every calendar month), the number of consecutive payments, the total value per range (and ideally if possible the gap between the current range and the end of the most recent one).

对于上面的测试数据，我的输出看起来像这样:

For the test data above my output would look like this:

CREATE TABLE #results
(
contact_reference NVARCHAR(55)
,Range_start DATETIME
,Range_end DATETIME
,Payments INT
,Value MONEY
,months_until_next_payment INT --works out the gap between the range_end date for a group and the range_start date for the next group
)
INSERT INTO #results VALUES('18EC3CD2-3065-4FF4-BE40-000004228590','2003-06-08','2003-06-08',1,12.82,12)
INSERT INTO #results VALUES('18EC3CD2-3065-4FF4-BE40-000004228590','2004-06-08','2004-06-08',1,12.82,6)
INSERT INTO #results VALUES('18EC3CD2-3065-4FF4-BE40-000004228590','2004-12-08','2004-12-08',1,12.82,4)
INSERT INTO #results VALUES('18EC3CD2-3065-4FF4-BE40-000004228590','2005-04-08','2006-02-28',12,153.843,2)
INSERT INTO #results VALUES('18EC3CD2-3065-4FF4-BE40-000004228590','2006-04-12','2008-06-06',27,416.6673,NULL)

我已经使用岛屿或迭代寻找答案，但坦率地说，我什至不知道从哪里开始将它们应用于我的问题，因此非常感谢任何帮助:)

I've looked for answers using islands, or iterations but I quite frankly don't even know where to begin applying them to my question, so any help massively appreciated :)

推荐答案

我已添加到 months_until_next_payment 列中.这会在应用程序中更有效地完成，而不是使用自联接，但是因为 SQL Server 没有任何特别令人满意的方式来引用下一行和上一行.

I've added in the months_until_next_payment column. This would be more efficiently done in the application rather than with a self join however as SQL Server does not have any particularly satisfactory way of referencing next and previous rows.

;WITH base AS ( 
SELECT    Contact_reference  ,
          Payment_value,
          DATEPART(YEAR, Date_payment)*12 + DATEPART(MONTH, Date_payment) - 
               DENSE_RANK() OVER 
                   (PARTITION BY Contact_reference 
                    ORDER BY DATEPART(YEAR, Date_payment)*12 + DATEPART(MONTH, Date_payment)) AS G,
          Date_payment
 FROM     #data
 ),
 cte AS
 (
 SELECT 
          Contact_reference, 
          ROW_NUMBER() over (partition by Contact_reference 
                                 order by MIN(Date_payment)) RN,
          MIN(Date_payment) Range_start,
          MAX(Date_payment) Range_end, 
          COUNT(Payment_value) Payments, 
          SUM(Payment_value) Value
 FROM base
 GROUP BY Contact_reference, G
 )
 SELECT 
       c1.Contact_reference, 
       c1.Payments, 
       c1.Range_end, 
       c1.Range_start, 
       c1.Value, 
       DATEDIFF(month, c1.Range_end,c2.Range_start) months_until_next_payment
 FROM cte c1
 LEFT join cte c2 ON c1.Contact_reference=c2.Contact_reference and c2.RN = c1.RN+1

这篇关于连续的日期的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！