Question

我尝试查找所有Cust成员，这些成员在2018年的每个月至少有一天的时间。

我想出了解决方案，例如在下面的代码段中，在每个月的开始/中/月末检查他们的成员资格，但是试图找到更智能的解决方案。

我知道我可以在365天的每一天内使用计数表进行检查，但是可能还有更优雅的解决方案？我对SQL有点陌生，我认为我在GROUPing领域中缺少了一些东西。

在下面显示的代码段中，两个Cust都有至少一天的会员资格。

所需的输出：

CustID
------
   1
  22

代码：

with data as 
(
    select * 
    from (values (1, 1,   '2017-12-11', '2018-01-16'),   (1, 22,  '2018-01-28', '2018-03-9' ), (1, 333, '2018-03-1', '2018-12-31') ,  -- island
                 (22, 1,  '2017-12-31', '2018-01-11'),   (22, 2,  '2017-2-11',  '2019-12-31')) as t (CustID, ContractID, StartDD, EndDD)     ---      
    )
    select 
        isdate(startDD), isdate(EndDD) 
    from 
        data
), gaps as 
(
    select  
        *,  
        datediff(day, lag(EndDD, 1, StartDD) over (partition by CustID order by StartDD), StartDD) as BreakDD          -- negative is island
    from 
        data
)
select 
    *, 
    datepart(month,StartDD) mmS , datepart(month,EndDD) mmE 
from 
    gaps 
    -- and was active any 1+ day during each of the 12 months in 2018    ????
where 
    1 = 1 
    /* and (cast('1/1/2018' as date) between StartDD and EndDD
            or cast('1/15/2018' as date) between StartDD and EndDD     
            or cast('1/31/2018' as date) between StartDD and EndDD)
       ---- etc..  for each month
       and (      cast('12/1/2018'  as date)   between  StartDD  and  EndDD 
              or  cast('12/15/2018' as date)   between  StartDD  and  EndDD  
              or  cast('12/31/2018' as date)   between  StartDD  and  EndDD  
           ) 
*/ 
--select CustID, max(BreakDD) Max_Days
--from gaps
--group by CustID

Answer 1

尝试这个答案。

首先创建一个函数，以返回给定日期之间的所有月份和年份。

功能

：

--SELECT * FROM dbo.Fn_GetMonthYear('2017-12-11','2018-01-16')
ALTER FUNCTION dbo.Fn_GetMonthYear(@StartDate  DATETIME,@EndDate    DATETIME)
RETURNS TABLE
AS

    RETURN(
    SELECT  DATEPART(MONTH, DATEADD(MONTH, x.number, @StartDate)) AS [Month]
            ,DATEPART(YEAR, DATEADD(MONTH, x.number, @StartDate)) AS [Year]
    FROM    master.dbo.spt_values x
    WHERE   x.type = 'P'        
    AND     x.number <= DATEDIFF(MONTH, @StartDate, @EndDate)
    )

表架构：

CREATE TABLE #t(CustID INT, ContractID INT, StartDD date, EndDD date)

INSERT INTO #t values (1, 1,   '2017-12-11', '2018-01-16'),   (1, 22,  '2018-01-28', '2018-03-9' ), (1, 333, '2018-03-1', '2018-12-31') ,  -- island
                 (22, 1,  '2017-12-31', '2018-01-11'),   (22, 2,  '2017-2-11',  '2019-12-31')

这是您需要的 T-SQL查询。

SELECT CustID
    ,COUNT(DISTINCT [Month]) NoOfMonths
FROM(
    SELECT * 
    FROM #t t
    CROSS APPLY dbo.Fn_GetMonthYear(StartDD,EndDD)
    )D
WHERE [Year] = 2018
GROUP BY CustID
HAVING COUNT(DISTINCT [Month])=12

结果：

CustID  NoOfMonths
1       12
22      12

Answer 2

查找所有拥有至少一天会员资格的Cust 2018年的一个月

我认为这意味着每个'2018-01-01'的数据必须存在于'2018-12-31'和custid之间。

CREATE TABLE #t(CustID INT, ContractID INT, StartDD date, EndDD date)

INSERT INTO #t values (1, 1,   '2017-12-11', '2018-01-16'),   (1, 22,  '2018-01-28', '2018-03-9' ), (1, 333, '2018-03-1', '2018-12-31') ,  -- island
 (22, 1,  '2017-12-31', '2018-01-11'),   (22, 2,  '2017-2-11',  '2019-12-31')

declare @From Datetime='2018-01-01'
declare @To datetime='2018-12-31'

;with CTE as
(
select CustID,min(StartDD)StartDD
,max(EndDD)EndDD
from #t
group by CustID
)
select CustID,StartDD
,EndDD
from CTE
where StartDD<=@From and EndDD>=@To

此脚本未在所有示例数据中进行测试。但是逻辑很明确，因此可以相应地进行纠正。

因此请告知哪些示例数据不起作用。

TSQL：全年/每月的连续时间

2 个答案: