我有一个问题,我感觉不到我的深度。
我需要在两个日期之间循环几个月,并返回每个月的数据子集,并且空白行数月,没有数据。
例如:
TransactionID | Date | Value
1 | 01/01/2015 | £10
2 | 16/01/2015 | £15
3 | 21/01/2015 | £5
4 | 15/03/2015 | £20
5 | 12/03/2015 | £15
6 | 23/04/2015 | £10
需要返回:
Month | Amount
January | £30
February | £0
March | £35
April | £10
我的查询将依赖于指定日期范围,以便我可以设置查询的第一个和最后一个日期。
我觉得我可能过度思考这个问题,但已经到了那个阶段,你开始觉得自己陷入了困境。
答案 0 :(得分:3)
关键是可以访问整数列表来表示范围内的月份。如果你没有Numbers Table,那么spt_values就会紧张。
SELECT
[Year] = YEAR(DATEADD(month,[i],@range_start))
,[Month] = DATENAME(month,DATEADD(month,[i],@range_start))
,[Amount] = ISNULL(SUM([Value]),0)
FROM (
SELECT TOP (DATEDIFF(month,@range_start,@range_end)+1)
ROW_NUMBER() OVER(ORDER BY (SELECT 1))-1 [i]
FROM master.dbo.spt_values
) t1
LEFT JOIN #MyTable t2
ON (t1.[i] = DATEDIFF(month,@range_start,t2.[Date]) )
GROUP BY [i]
ORDER BY [i]
答案 1 :(得分:2)
SQL起初是一种棘手的语言。你实际上不想要一个循环。实际上,除了极少数情况外,你几乎不想在SQL中循环。试试这个:
DECLARE @StartDate DATE,
@EndDate DATE;
SET @StartDate = '01 January 2015';
SET @EndDate = '30 April 2015';
WITH CTE_Months
AS
(
SELECT @StartDate dates
UNION ALL
SELECT DATEADD(MONTH,1,dates)
FROM CTE_Months
WHERE DATEADD(MONTH,1,dates) < @EndDate
)
SELECT YEAR(B.[date]) AS yr,
DATENAME(MONTH,B.[Date]) AS month_name,
SUM(ISNULL(B.Value,0)) AS Amount
FROM CTE_Months A
LEFT JOIN yourTable B
ON YEAR(A.[date]) = YEAR(B.[date])
AND MONTH(A.[date]) = MONTH(B.[date])
GROUP BY YEAR(B.[date]),DATENAME(MONTH,B.[Date])
答案 2 :(得分:0)
一种方法:创建一个名为months
的表格,其中包含monthnum
int
字段和12行[1..12]
declare @start date = '01 jan 2015',
@end date = '30 apr 2015'
select
datename(month, dateadd(month, monthnum, 0) - 1),
isnull(Amount, 0)
from months
left join (
select
month(date) Month,
sum(Value) Amount
from tbl
where date between @start and @end
group by month(date)
) T on (T.Month = months.monthnum)
where months.monthnum between month(@start) and month(@end)
order by monthnum
答案 3 :(得分:0)
The following code will generate one output row for each month between the first and last transaction dates. Spanning a year boundary, or multiple years, is handled correctly.
[link]