多列重叠日期-Java 学习之路

;with cte as (
 select Domain_Id, Starting_Date, End_Date
 from Que_Date
 union all
 select t.Domain_Id, cte.Starting_Date, t.End_Date
 from cte
 join Que_Date t on cte.Domain_Id = t.Domain_Id and cte.End_Date = t.Starting_Date),
cte2 as (
 select *, rn = row_number() over (partition by Domain_Id, End_Date order by Domain_Id)
 from cte
)
select DISTINCT Domain_Id, Starting_Date, max(End_Date) enddate
from cte2
where rn=1
group by Domain_Id, Starting_Date
order by Domain_Id, Starting_Date;

select * from Que_Date

这是我写的代码，但是我得到了一个额外的行，即第二行是额外的，预期的输出应该只有第1行，第3行和第4行作为输出，所以请帮助我 .

我附上了一个显示输入，例外输出和我得到的输出的图像 .

2 回答

我使用了@stepup提供的相同的表创建脚本，但您也可以通过这种方式获得结果 .

CREATE TABLE testtbl
(
Domain_ID INT,
Starting_Date DATE,
End_Date Date
)

INSERT INTO testtbl

VALUES
(1, convert(date, '01-01-2011' ,103), convert(date, '05-01-2011',103)  )
,(1,  convert(date, '05-01-2011' ,103), convert(date, '07-01-2011',103)  )  
,(1,  convert(date, '07-01-2011' ,103), convert(date, '15-01-2011',103)  )
,(2, convert(date, '11-05-2011' ,103), convert(date, '12-05-2011',103)   )
,(2, convert(date, '13-05-2011' ,103), convert(date, '14-05-2011',103)   )

您可以使用自联接和Firs_value以及组中的最后一个值，以确保您在相同的ID和重叠日期内进行比较 .

select distinct t.Domain_ID,
case when lag(t1.starting_date)over (partition by t.Domain_id order by 
t.starting_date) is not null   
then first_value(t.Starting_Date) over (partition by t.domain_id order by 
t.starting_date) 
else  t.Starting_Date end StartingDate, 
case when lead(t.domain_id) over (partition by t.domain_id order by t.starting_date)  = 
t1.Domain_ID then  isnull(last_value(t.End_Date) over (partition by t.domain_id  order by t.end_date rows between unbounded preceding and unbounded following),t.End_Date)   
else t.End_Date  end    end_date 
from testtbl t 
left join testtbl t1 on t.Domain_ID = t1.Domain_ID 
                    and t.End_Date = t1.Starting_Date 
                    and t.Starting_Date < t1.Starting_Date

输出：

Domain_ID   StartingDate    end_date
1           2011-01-01  2011-01-15
2           2011-05-11  2011-05-12
2           2011-05-13  2011-05-14

回复于 2024-04-20T07:45:05+08:00

你的第一个 cte 你有这么多的结果 . 您的第一个 cte 包含域名 . 因此，您无法根据 cte 过滤域名 . 所以你查询有不必要的行 .

尝试此解决方案 . Cte ConsistentDomains 只有一致的域名 . 因此，基于此cte，我们可以得到不重叠的结果 .

创建和填充数据：

CREATE TABLE FooTable
(
    Domain_ID INT,
    Starting_Date DATE,
    End_Date Date
)

INSERT INTO dbo.FooTable
(
    Domain_ID,
    Starting_Date,
    End_Date
)
VALUES
(   1,         -- Domain_ID - int
    CONVERT(datetime,'01-01-2011',103), -- Starting_Date - date
    CONVERT(datetime,'05-01-2011',103)  -- End_Date - date
    )
, (1, CONVERT(datetime,'05-01-2011',103), CONVERT(datetime,'07-01-2011',103))
, (1, CONVERT(datetime,'07-01-2011',103), CONVERT(datetime,'15-01-2011',103))
, (2, CONVERT(datetime,'11-05-2011',103), CONVERT(datetime,'12-05-2011',103))
, (2, CONVERT(datetime,'13-05-2011',103), CONVERT(datetime,'14-05-2011',103))

查询以查找不重叠的结果：

DECLARE @startDate varchar(50) = '2011-01-01';
WITH ConsistentDomains AS
(
    SELECT      
      f.Domain_ID
    , f.Starting_Date
    , f.End_Date    
    FROM FooTable f
    WHERE f.Starting_Date = @startDate
    UNION ALL
    SELECT 
      s.Domain_ID
    , s.Starting_Date
    , s.End_Date
    FROM FooTable s
    INNER JOIN  ConsistentDomains cd 
      ON s.Domain_ID = cd.Domain_ID
      AND s.Starting_Date = cd.End_Date
), ConsistentDomainsRownumber AS 
(
    SELECT 
      cd.Domain_ID
    , cd.Starting_Date
    , cd.End_Date
    , ROW_NUMBER() OVER (PARTITION BY cd.Domain_ID ORDER BY cd.Starting_Date, 
        cd.End_Date) RN
    FROM ConsistentDomains cd
)
SELECT cd.Domain_ID
, convert(varchar, cd.Starting_Date, 105) Starting_Date
, convert(varchar, cd.End_Date, 105) End_Date
FROM ConsistentDomainsRownumber cd WHERE  cd.RN = 1
UNION ALL
SELECT        
   ft.Domain_ID
   , convert(varchar, ft.Starting_Date, 105) Starting_Date
   , convert(varchar, ft.End_Date, 105) End_Date
 FROM dbo.FooTable ft WHERE ft.Domain_ID NOT IN (SELECT cd.Domain_ID FROM 
     ConsistentDomainsRownumber cd)

输出：

回复于 2024-04-20T07:45:05+08:00

多列重叠日期

2 回答

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