我有两个表,每个表都保存日期范围(从date1到date2)
我将在表1和表2中的两个日期之间找到重叠的天
示例
table1
-------------------------
id | FromDate | ToDate
1 |2000-01-01 | 2000-02-04
2 |2000-03-01 | 2000-03-29
table2
-------------------------
id | FromDate | ToDate
1 |2000-02-01 | 2000-02-07
2 |2000-03-27 | 2000-03-29
我想要的结果:
2000-02-01
2000-02-02
2000-02-03
2000-02-04
2000-03-27
2000-03-28
2000-03-29
答案 0 :(得分:1)
这应该有效:
CREATE TABLE #t1
(
id int,
FromDate date,
ToDate date
)
CREATE TABLE #t2
(
id int,
FromDate date,
ToDate date
)
INSERT #t1 VALUES
(1, '2000-01-01', '2000-02-04'),
(2, '2000-03-01', '2000-03-29')
INSERT #t2 VALUES
(1, '2000-02-01', '2000-02-07'),
(2, '2000-03-27', '2000-03-29')
WITH DateRange AS --select range where intersection is possible
(
SELECT MAX(MinDate) MinDate,MIN(MaxDate) MaxDate,DATEDIFF(DAY,MAX(MinDate),MIN(MaxDate)) Diff
FROM (VALUES ((SELECT MIN(FromDate) FROM #t1)),((SELECT MIN(FromDate) FROM #t2))) MinDate(MinDate)
CROSS APPLY (VALUES ((SELECT MAX(ToDate) FROM #t1)),((SELECT MAX(ToDate) FROM #t2))) MaxDate(MaxDate)
), AllDates AS --generate sequence of days
(
SELECT MinDate D, MaxDate Limit
FROM DateRange
UNION ALL
SELECT DATEADD(DAY, 1, D), Limit
FROM AllDates
WHERE DATEADD(DAY, 1, D)<=Limit
) --select all days existing in any range in both tables
SELECT D
FROM AllDates
WHERE EXISTS (SELECT * FROM #t1 WHERE D>=FromDate AND D<=ToDate)
AND EXISTS (SELECT * FROM #t2 WHERE D>=FromDate AND D<=ToDate)
答案 1 :(得分:1)
可以通过CTE和递归来做到这一点。
--Your sample data
DECLARE @table1 TABLE (id int PRIMARY KEY, FromDate date, ToDate date)
DECLARE @table2 TABLE (id int PRIMARY KEY, FromDate date, ToDate date)
INSERT INTO @table1 VALUES (1, '2000-01-01', '2000-02-04') , (2, '2000-03-01', '2000-03-29')
INSERT INTO @table2 VALUES (1, '2000-02-01', '2000-02-07') , (2, '2000-03-27', '2000-03-29')
--A couple CTE's
;WITH cteDates AS (
SELECT T1.id --get the min and max dates for each id
,CASE WHEN T1.FromDate > T2.FromDate THEN T1.FromDate ELSE T2.FromDate END [mindate]
,CASE WHEN T1.ToDate < T2.ToDate THEN T1.ToDate ELSE T2.ToDate END [maxdate]
FROM @table1 T1 INNER JOIN @table2 T2 ON T1.id = T2.id
)
, cteRecursion AS ( --date range for each id
SELECT id, mindate AS DateValue
FROM cteDates
UNION ALL
SELECT id, DATEADD(DAY, 1, DateValue)
FROM cteRecursion C1
WHERE DATEADD(DAY, 1, DateValue) <= (
SELECT maxDate
FROM cteDates C2
WHERE C2.id = C1.id
)
)
--SELECT query
SELECT DateValue FROM cteRecursion ORDER BY DateValue OPTION (MAXRECURSION 0)
产生输出:
DateValue
---------
2000-02-01
2000-02-02
2000-02-03
2000-02-04
2000-03-27
2000-03-28
2000-03-29
答案 2 :(得分:0)
一种可能的解决方案是使用数字或理货价格表
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结果是
;WITH cteNumbers (N)
AS(
SELECT ROW_NUMBER() OVER(ORDER BY N1.N)
FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N1(N)
CROSS JOIN (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N2 (N)
CROSS JOIN (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N3 (N)
)
SELECT T1.FromDate
FROM(
SELECT
T1.FromDate
FROM dbo.Table1 T1
UNION
SELECT
DATEADD(DAY, N, T1.FromDate)
FROM
dbo.Table1 T1
CROSS APPLY cteNumbers N
WHERE N <= DATEDIFF(DAY, T1.FromDate, T1.ToDate)
) T1
WHERE t1.FromDate IN
(
SELECT
T2.FromDate
FROM dbo.Table2 T2
UNION
SELECT
DATEADD(DAY, N, T2.FromDate)
FROM
dbo.Table2 T2
CROSS APPLY cteNumbers N
WHERE N <= DATEDIFF(DAY, T2.FromDate, T2.ToDate)
)
“数字/总计”表允许的日期范围最长为1000天。如果您需要更多,请添加另一行,例如FromDate
2000-02-01 00:00:00.000
2000-02-02 00:00:00.000
2000-02-03 00:00:00.000
2000-02-04 00:00:00.000
2000-03-27 00:00:00.000
2000-03-28 00:00:00.000
2000-03-29 00:00:00.000