Question

我有以下SQL，我希望在以下日期有差距。

declare @startdate datetime = '2017-05-01'
declare @enddate datetime = '2017-05-25'

create table #tmpdates (id int, date1 datetime, date2 datetime, rate int)

insert into #tmpdates values (1, '2017-05-05', '2017-05-15', 10)
insert into #tmpdates values (2, '2017-05-16', '2017-05-18', 12)
insert into #tmpdates values (3, '2017-05-21', '2017-05-25', 15)

select * from #tmpdates where date1 >= @startdate and date2 <= @enddate

drop table #tmpdates

因此输出应包含2017-05-01至2017-05-04和2017-05-19至2017-05-20 - 另外2条记录。

Output:
1   5/1/2017 0:00   5/4/2017 0:00   NO DATA
2   5/5/2017 0:00   5/15/2017 0:00  10
3   5/16/2017 0:00  5/18/2017 0:00  12
4   5/19/2017 0:00  5/20/2017 0:00  NO DATA
5   5/21/2017 0:00  5/25/2017 0:00  15

在我的上述查询中，只有日期范围记录返回..请指导或如何包含这些？

Answer 1

这可以假设没有重叠的间隔。

declare @startdate datetime = '2017-05-16'
declare @enddate datetime = '2017-05-26'

create table #tmpdates (id int, date1 datetime, date2 datetime, rate int)

insert into #tmpdates values (0, '2017-04-01', '2017-04-25',22)
insert into #tmpdates values (1, '2017-05-05', '2017-05-15', 10)
insert into #tmpdates values (2, '2017-05-16', '2017-05-18', 12)
insert into #tmpdates values (3, '2017-05-21', '2017-05-25', 15)

declare @final_result table (date1 date, date2 date, rate int)

insert into @final_result 

select @startdate,dateadd(day,-1,t.date1),null
from #tmpdates t
where @startdate < t.date1 and 
        t.date1 <= (select min(t1.date1) from #tmpdates t1 where t1.date1 >= @startdate)

union all

select date1, date2, rate 
from #tmpdates 
where (date1 >= @startdate or date2 >= @startdate) and 
      (date2 <= @enddate or date1 <= @enddate)

union all

select dateadd(day,1,t.date2), 
        ( select dateadd(day,-1,min(t3.date1)) 
            from #tmpdates t3 where t3.date1 > t.date2) , 
        null
from #tmpdates t
where dateadd(day,1,t.date2) < (select min(t1.date1) from #tmpdates  t1 where t1.date1 > t.date2)
and t.date1 >= @startdate and t.date2 <= @enddate

union all

select dateadd(day,1,max(t.date2)), @enddate, null
from #tmpdates t
having max(t.date2) < @enddate


drop table #tmpdates

select * from @final_result order by date1

修改

它从四个查询中收集数据并执行union all。

第一个查询：

select @startdate,dateadd(day,-1,t.date1),null from #tmpdates t where @startdate < t.date1 and t.date1 <= (select min(t1.date1) from #tmpdates t1 where t1.date1 >= @startdate)

如果在忽略@startdate之前有间隔，则选择表中@startdate和第一个（最小）日期之间的差距。因此，它会选择从@startdate到大于@startdate的时间间隔的第一个日期的差距（如果有的话）。

第二个查询：

select date1, date2, rate from #tmpdates where (date1 >= @startdate or date2 >= @startdate) and (date2 <= @enddate or date1 <= @enddate)

从表中选择记录（非间隙）。如果@startdate介于范围之间，则包含该记录。 @enddate参数也是如此。

第三个查询：

select dateadd(day,1,t.date2), ( select dateadd(day,-1,min(t3.date1)) from #tmpdates t3 where t3.date1 > t.date2) , null from #tmpdates t where dateadd(day,1,t.date2) < (select min(t1.date1) from #tmpdates t1 where t1.date1 > t.date2) and t.date1 >= @startdate and t.date2 <= @enddate

选择表格中最小和最大（介于@startdate和@enddate之间）间隔之间的差距。

最后 第四个查询：

select dateadd(day,1,max(t.date2)), @enddate, null from #tmpdates t having max(t.date2) < @enddate

选择表格中最大日期（@startdate和@enddate之间的最大日期）与@enddate之间的差距，如果有差距的话。

所有这些记录都插入到@final_result表中，以便按间隔排序。

Answer 2

请使用以下查询：

DECLARE @STARTDATE DATE = '2017-05-01'
DECLARE @ENDDATE DATE = '2017-05-25'

DECLARE @DATES TABLE (ID INT, DATE1 DATE, DATE2 DATE, RATE INT)

INSERT INTO @DATES VALUES 
(1, '2017-05-05', '2017-05-15', 10),
(2, '2017-05-16', '2017-05-19', 12),
(3, '2017-05-21', '2017-05-25', 15)

SELECT* FROM 
(
    SELECT @STARTDATE AS DATE1,DATEADD(DAY,-1,MIN(DATE1)) AS DATE2,'NO DATA'AS RATE FROM @DATES
    UNION
    SELECT 
    CASE WHEN   
            LEAD(DATE1) OVER (ORDER BY DATE1) = DATEADD(DAY,1,DATE2) THEN NULL 
            ELSE DATEADD(DAY,1,DATE2) END AS DATE1,
    CASE WHEN   
            LEAD(DATE1) OVER (ORDER BY DATE1) = DATEADD(DAY,1,DATE2) THEN NULL 
            ELSE LEAD(DATEADD(DAY,-1,DATE1)) OVER (ORDER BY DATE1) END AS DATE2,
    'NO DATA'AS RATE
    FROM @DATES d
    UNION
    SELECT DATE1, DATE2,CAST(RATE AS NVARCHAR(10)) FROM @DATES
    UNION
    SELECT DATEADD(DAY,1,MAX(DATE2)) AS DATE1,@ENDDATE AS DATE2,'NO DATA'AS RATE FROM @DATES
) A WHERE A.DATE2 IS NOT NULL AND A.DATE1 <= A.DATE2
  AND DATE1 >= @STARTDATE AND DATE2 <=@ENDDATE
ORDER BY A.DATE1

Answer 3

您可以尝试以下代码。我正在从@StartDate走到@endDate并找到差距。

declare @startdate datetime = '2017-05-01'
declare @enddate datetime = '2017-05-04'
declare @startdate1 datetime, @enddate1 datetime
declare @dates table (date1 date,date2 date)
create table #tmpdates (id int, date1 datetime, date2 datetime, rate int)

insert into #tmpdates values (1, '2017-05-05', '2017-05-15', 10)
insert into #tmpdates values (2, '2017-05-16', '2017-05-18', 12)
insert into #tmpdates values (3, '2017-05-21', '2017-05-25', 15)

select * from #tmpdates where date1 >= @startdate and date2 <= @enddate
set @startdate1=@startdate
while @startdate1<=@enddate
begin
     if not exists(select 1 from #tmpdates where @startdate1 between date1 and date2)
     begin
          if not exists (select 1 from @dates where @startdate1 > date1 and date2 is null)
          begin
               insert into @dates(date1)values(@startdate1)
          end
          else
          begin
               if @startdate1+1>=@enddate
               begin
                   update @dates set date2=@startdate1 where date2 is null
               end
               set @startdate1+=1
          end
     end
     else
     begin 
         update @dates set date2=@startdate1-2 where date2 is null
     end
     set @startdate1+=1       
end
select * from
(select date1,date2, rate from #tmpdates
union
select *,0  as rate from @dates
) A WHERE date1>=@startdate and date2<=@enddate
drop table #tmpdates

查找日期范围中的差距 - TSQL

3 个答案: