寻找连续 4 年捐赠的客户（包括差距）

Question

我有一个 table 类似于以下内容：

+------------+-----------+
| CustomerID | OrderYear |
+------------+-----------+
|          1 |      2012 |
|          1 |      2013 |
|          1 |      2014 |
|          1 |      2017 |
|          1 |      2018 |
|          2 |      2012 |
|          2 |      2013 |
|          2 |      2014 |
|          2 |      2015 |
|          2 |      2017 |
+------------+-----------+

我如何确定哪些 CustomerID 连续 4 年捐赠？（上面只有客户2。）可以看到，有些记录在顺序年份会有差距。

我开始尝试使用 ROW_NUMBER/LAG/LEAD 的某种组合，但到目前为止运气不佳。

非常配对 down/modified 尝试...

WITH CTE
AS
(
SELECT T.ConstituentLookupID,
       T.FISCALYEAR,
       COUNT(T.FISCALYEAR) OVER (PARTITION BY T.ConstituentLookupID) AS 
YearCount,
   FIRST_VALUE(T.FISCALYEAR) OVER(PARTITION BY T.ConstituentLookupID ORDER 
BY T.FISCALYEAR DESC) - T.FISCALYEAR + 1 as X,
   ROW_NUMBER() OVER(PARTITION BY T.ConstituentLookupID ORDER BY 
T.FISCALYEAR DESC) AS RN
FROM #Temp AS T)
SELECT CTE.ConstituentLookupID,
   CTE.FISCALYEAR,
   CTE.YearCount,
   CTE.X,
   CTE.RN,
FROM CTE
WHERE CTE.YearCount >= 4 --Have at least 4 years of giving
AND CTE.X - CTE.RN = 1 --Some kind of way to calculate consecutive years. Doesnt account current year and gaps...;

Answer 1

假设没有重复，你可以使用lag():

select distinct customerid
from (
    select t.*, 
        lag(orderyear, 3) over(partition by customerid order by orderyear) oderyear3
    from mytable t
) t
where orderyear = orderyear3 + 3

一种更传统的方法是使用一些间隙和孤岛技术。如果您想要每个系列的开始和结束，这很方便。在这里，一个岛是一系列具有“相邻”顺序年份的行，并且您需要至少 4 年长的岛。我们可以通过将顺序年份与递增序列进行比较来识别岛屿，然后使用聚合：

select customerid, min(orderyear) firstorderyear, max(orderyear) lastorderyear
from (
    select t.*, 
        row_number() over(partition by customerid order by orderyear) rn
    from mytable t
) t
group by customerid, orderyear - rn
having count(*) >= 4

Answer 2

我有一个使用行号和分组依据的简单解决方案

SELECT Max(z.customerid), 
       Count(z.grp) 
FROM   (SELECT customerid, 
               orderyear, 
               orderyear - Row_number() 
                             OVER ( 
                               ORDER BY customerid) AS Grp 
        FROM   mytable)z 
GROUP  BY z.grp 
HAVING Count(z.grp) = 4

Answer 3

假设每个客户和年份不超过一行，最简单的方法是 lag():

select customerid, year
from (select t.*,
             lag(orderyear, 3) over (partition by customerid order by orderyear) as prev3_year
      from t
     ) t
where prev3_year = year - 3;

这个想法是回顾 3 年前。如果那一年是第 3 年，则连续四年。如果您的数据可以有重复项，则可以对逻辑进行调整（它们会使查询稍微复杂一些）。

这可能 return 重复，所以您可能只想：

select distinct customerid
from (select t.*,
             lag(orderyear, 3) over (partition by customerid order by orderyear) as prev3_year
      from t
     ) t
where prev3_year = year - 3;

寻找连续 4 年捐赠的客户（包括差距）

Find Customers With 4 Consecutive Years of Giving (Including Gaps)

sql

tsql

sql-server

count

gaps-and-islands