重叠日期范围 - 仅识别重叠

Question

我已经看到很多解决方案来识别日期范围重叠的记录，还有其他合并重叠范围的示例。

但是我对显示只有重叠发生的范围的结果感兴趣。事实上，我有3个ProductID（并且只有3个将存在），并且我试图找出每个客户的全部三个日期范围。

SET NOCOUNT ON; 

CREATE TABLE #tmp 
(
    CustomerID integer 
    ,ProductID varchar(12) 
    ,Eff_Dt DATE 
    ,End_Dt DATE 
); 

-- Customer 1000: Expecting results to show 2 rows: 1/1 - 1/5 and 1/10 - 1/15 
INSERT INTO #tmp VALUES (1000,'PRODUCT_A','01-01-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (1000,'PRODUCT_B','01-01-2013' ,'01-05-2013'); 
INSERT INTO #tmp VALUES (1000,'PRODUCT_B','01-10-2013' ,'01-15-2013'); 
INSERT INTO #tmp VALUES (1000,'PRODUCT_C','01-01-2013' ,'01-31-2013'); 

-- Customer 2000: Expecting results to show 1 row: 1/19 - 1/31 
INSERT INTO #tmp VALUES (2000,'PRODUCT_A','01-01-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (2000,'PRODUCT_B','01-01-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (2000,'PRODUCT_C','01-19-2013' ,'01-31-2013'); 

-- Customer 3000: Expecting results to show no rows (or nulls) 
INSERT INTO #tmp VALUES (3000,'PRODUCT_A','01-01-2013' ,'01-10-2013'); 
INSERT INTO #tmp VALUES (3000,'PRODUCT_A','01-16-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (3000,'PRODUCT_B','01-01-2013' ,'01-12-2013'); 
INSERT INTO #tmp VALUES (3000,'PRODUCT_C','01-15-2013' ,'01-31-2013'); 

-- Customer 4000: Expecting results to show 1 row: 1/15 - 1/23 
INSERT INTO #tmp VALUES (4000,'PRODUCT_A','01-15-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (4000,'PRODUCT_B','01-01-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (4000,'PRODUCT_C','01-01-2013' ,'01-23-2013'); 

-- Customer 5000: Expecting results to show 0 rows 
INSERT INTO #tmp VALUES (5000,'PRODUCT_A','01-17-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (5000,'PRODUCT_B','01-01-2013' ,'01-10-2013'); 
INSERT INTO #tmp VALUES (5000,'PRODUCT_C','01-07-2013' ,'01-19-2013'); 

-- Customer 6000: Expecting results to show 3 rows: 1/11 - 1/12 1/17 - 1/22 1/26 - 1/27 
INSERT INTO #tmp VALUES (6000,'PRODUCT_A','01-01-2013' ,'01-04-2013'); 
INSERT INTO #tmp VALUES (6000,'PRODUCT_A','01-09-2013' ,'01-12-2013'); 
INSERT INTO #tmp VALUES (6000,'PRODUCT_A','01-17-2013' ,'01-22-2013'); 
INSERT INTO #tmp VALUES (6000,'PRODUCT_A','01-26-2013' ,'01-31-2013'); 
INSERT INTO #tmp VALUES (6000,'PRODUCT_B','01-04-2013' ,'01-28-2013'); 
INSERT INTO #tmp VALUES (6000,'PRODUCT_C','01-11-2013' ,'01-27-2013'); 

SET NOCOUNT OFF; 
/* ====== EXPECTED RESULTS ======================= 

CustomerID EFF_DT  END_DT 
1000   1/1/2013  1/5/2013 
1000   1/10/2013  1/15/2013 
2000   1/19/2013  1/31/2013 
4000   1/15/2013  1/23/2013 
6000   1/11/2013  1/12/2013 
6000   1/17/2013  1/22/2013 
6000   1/26/2013  1/27/2013 

===================================================*/ 

Answer 1

这里的答案：

select t.customerid, t.eff_dt, count(distinct t2.productId), 
     MIN(t2.end_dt) as end_dt 
from #tmp t join 
    #tmp t2 
    on t.CustomerID = t2.CustomerID and 
     t.Eff_Dt between t2.Eff_Dt and t2.End_Dt 
group by t.CustomerID, t.eff_dt 
having count(distinct t2.productId) = 3 

这是使用自联接来计算不同产品的每个eff_dt数量。你需要三种不同的产品，这就是having条款正在做的事情。

有三种不同的产品，直到其中一个结束。这将是生效日期后的第一个end_dt - 由min(end_dt)计算。

Answer 2

尝试：

select ab.CustomerID, 
     case when ab_Eff_Dt > c.Eff_Dt then ab_Eff_Dt else c.Eff_Dt end abc_Eff_Dt, 
     case when ab_End_Dt < c.End_Dt then ab_End_Dt else c.End_Dt end abc_End_Dt 
from 
(select a.CustomerID, 
     case when a.Eff_Dt > b.Eff_Dt then a.Eff_Dt else b.Eff_Dt end ab_Eff_Dt, 
     case when a.End_Dt < b.End_Dt then a.End_Dt else b.End_Dt end ab_End_Dt 
from #tmp a 
join #tmp b 
    on a.CustomerID = b.CustomerID and a.Eff_Dt < b.End_Dt and b.Eff_Dt < a.End_Dt 
where a.ProductID = 'PRODUCT_A' and b.ProductID = 'PRODUCT_B') ab 
join #tmp c 
    on ab.CustomerID = c.CustomerID and ab_Eff_Dt < c.End_Dt and c.Eff_Dt < ab_End_Dt 
where c.ProductID = 'PRODUCT_C' 

（SQLFiddle here）