差距和岛屿？或不？

Question

我有以下 table:

;with data as (
select '508325'  as [customer], 'G61' as [demo_given],
        cast('2015-1-1' as date) as [date_of_demo]
union all select '508325', 'G61', cast('2015-3-1' as date) 
union all select '508325', 'G61',cast('2015-3-15' as date)
union all select '508325', 'G61',cast('2015-3-16' as date)
union all select '508325', 'G61',cast('2015-3-17' as date)
union all select '508325', 'G61',cast('2015-6-1' as date)
union all select '508325', 'G61',cast('2015-8-1' as date)
union all select '508325', 'G61',cast('2015-9-1' as date)
union all select '508325', 'G61',cast('2015-9-1' as date)
union all select '508325', 'G61',cast('2015-12-1' as date)
)

每个客户在 4 个月内只能进行 3 次演示。第一个时期从给出的第一个演示开始计算，并在 4 个月后结束。

如果那段时间的演示数量超过 3，我需要那 4 个月期间的演示 4 和之后的日期。（在本例中为 2015-3-16 和 2015-3-17）

下一期从前四个月后第一次演示的日期开始。因此，我需要计算 2015-6-1 期间的演示数量，直到 2015-9-30 和 return 期间最终 'surplus' 演示的日期。

我该怎么做？

Answer 1

为了便于阅读，我使用了多步 CTE，但如果需要，您可以将其组合：

tally - 简单数字 table 你可以使用任何你想要的方法（递归 cte，虚拟 table，table 函数，...）
min_date_per_customer - 获取每个客户的首次演示日期
date_ranges - 生成范围将 4 个月添加到 min_date
final - 加入 data 到 date_ranges，生成行号
main query - 过滤掉特定时期内第 4、5、6、... 的演示

代码：

WITH tally(N) AS (
  SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
  FROM sys.all_columns a CROSS JOIN sys.all_columns b
), min_date_per_customer AS (
  SELECT customer, MIN(date_of_demo) AS min_date
  FROM #data
  GROUP BY customer
), date_ranges AS (
  SELECT t.N, mdpc.customer
      ,[date_start] = DATEADD(m, 4 * (t.N - 1), min_date)
      ,[date_end]   = DATEADD(m, 4 *t.N, min_date)
  FROM min_date_per_customer mdpc
  CROSS JOIN tally t
  WHERE t.N < 100 -- you can generate as many period as you wish
), final AS (
  SELECT d.customer
       ,d.demo_given
       ,d.date_of_demo
       ,dr.N
       ,rn = ROW_NUMBER() OVER (PARTITION BY dr.customer, dr.N ORDER BY date_of_demo)
  FROM #data d
  JOIN date_ranges dr
    ON d.[date_of_demo] >= dr.date_start
   AND d.[date_of_demo] <= dr.date_end
   AND d.customer = dr.customer
)
SELECT *
FROM final
WHERE rn > 3
ORDER BY customer, date_of_demo;

LiveDemo

输出：

╔══════════╦════════════╦═════════════════════╦═══╦═════╗
║ customer ║ demo_given ║    date_of_demo     ║ N ║ rn  ║
╠══════════╬════════════╬═════════════════════╬═══╬═════╣
║   508325 ║ G61        ║ 2015-03-16 00:00:00 ║ 1 ║   4 ║
║   508325 ║ G61        ║ 2015-03-17 00:00:00 ║ 1 ║   5 ║
║   508325 ║ G61        ║ 2015-09-01 00:00:00 ║ 2 ║   4 ║
╚══════════╩════════════╩═════════════════════╩═══╩═════╝

差距和岛屿？或不？

Gaps and Islands ? Or Not?

tsql

sql-server

gaps-and-islands

sql-server-2012