declare @MaxDate date
declare @MinDate date
select @MaxDate = MAX([Date]),
@MinDate = MIN([Date])
from Dates
declare @MaxValue int
declare @MinValue int
select @MaxValue = [Value] from Dates where [Date] = @MaxDate
select @MinValue = [Value] from Dates where [Date] = @MinDate
declare @diff int
select @diff = DATEDIFF(d, @MinDate, @MaxDate)
declare @increment int
set @increment = (@MaxValue - @MinValue)/@diff
select @increment
declare @jaggedDates as table
(
PID INT IDENTITY(1,1) PRIMARY KEY,
ThisDate date,
ThisValue int
)
declare @finalDates as table
(
PID INT IDENTITY(1,1) PRIMARY KEY,
[Date] date,
Value int
)
declare @thisDate date
declare @thisValue int
declare @nextDate date
declare @nextValue int
declare @count int
insert @jaggedDates select [Date], [Value] from Dates
select @count = @@ROWCOUNT
declare @thisId int
set @thisId = 1
declare @entryDiff int
declare @missingDate date
declare @missingValue int
while @thisId <= @count
begin
select @thisDate = ThisDate,
@thisValue = ThisValue
from @jaggedDates
where PID = @thisId
insert @finalDates values (@thisDate, @thisValue)
if @thisId < @count
begin
select @nextDate = ThisDate,
@nextValue = ThisValue
from @jaggedDates
where PID = @thisId + 1
select @entryDiff = DATEDIFF(d, @thisDate, @nextDate)
if @entryDiff > 1
begin
set @missingDate = @thisDate
set @missingValue = @thisValue
while @entryDiff > 1
begin
set @missingDate = DATEADD(d, 1, @missingDate)
set @missingValue = @missingValue + @increment
insert @finalDates values (@missingDate, @missingValue)
set @entryDiff = @entryDiff - 1
end
end
end
set @thisId = @thisId + 1
end
select * from @finalDates
感謝GalacticJello你是明星。就是我以後的樣子。 – SausageFingers 2010-06-10 22:43:42
該解決方案基於表中的第一個和最後一個條目計算因子的缺失值。我稍微修改了代碼以重新計算基於任何給定行的前一個和下一個已知值的缺失值。在該行之後: 「其中PID = @thisId + 1」 添加行: 「select @diff = DATEDIFF(d,@thisDate,@nextDate)」 「set @increment =(@nextValue - @thisValue)/ @diff」 – SausageFingers 2010-06-13 17:41:43