这里是一个快速的解决方案(可能不是最有效的):
SQL> CREATE TABLE myData AS
2 SELECT 'A' name, date'2010-01-01' d1, date'2010-12-11' d2 FROM DUAL
3 UNION ALL SELECT 'B', date'2010-01-20', date'2010-04-15' FROM DUAL
4 UNION ALL SELECT 'B', date'2010-05-10', date'2010-12-30' FROM DUAL
5 UNION ALL SELECT 'C', date'2010-03-13', date'2010-06-10' FROM DUAL;
Table created
SQL> WITH segments AS (
2 SELECT dat seg_low, lead(dat) over(ORDER BY dat) seg_high
3 FROM (SELECT d1 dat FROM myData
4 UNION
5 SELECT d2 dat FROM myData)
6 )
7 SELECT s.seg_low, s.seg_high
8 FROM segments s
9 JOIN myData m ON s.seg_high > m.d1
10 AND s.seg_low < m.d2
11 GROUP BY s.seg_low, s.seg_high
12 HAVING COUNT(DISTINCT NAME) = 3;
SEG_LOW SEG_HIGH
----------- -----------
13/03/2010 15/04/2010
10/05/2010 10/06/2010
我建立所有可能的连续日期范围,加入这个“日历”与样本数据。这将列出所有具有3个值的范围。您可能需要合并的结果,如果你添加行:
SQL> insert into mydata values ('B',date'2010-04-15',date'2010-04-16');
1 row inserted
SQL> WITH segments AS (
2 SELECT dat seg_low, lead(dat) over(ORDER BY dat) seg_high
3 FROM (SELECT d1 dat FROM myData
4 UNION
5 SELECT d2 dat FROM myData)
6 )
7 SELECT MIN(seg_low), MAX(seg_high)
8 FROM (SELECT seg_low, seg_high, SUM(gap) over(ORDER BY seg_low) grp
9 FROM (SELECT s.seg_low, s.seg_high,
10 CASE
11 WHEN s.seg_low
12 = lag(s.seg_high) over(ORDER BY s.seg_low)
13 THEN 0
14 ELSE 1
15 END gap
16 FROM segments s
17 JOIN myData m ON s.seg_high > m.d1
18 AND s.seg_low < m.d2
19 GROUP BY s.seg_low, s.seg_high
20 HAVING COUNT(DISTINCT NAME) = 3))
21 GROUP BY grp;
MIN(SEG_LOW) MAX(SEG_HIGH)
------------ -------------
13/03/2010 16/04/2010
10/05/2010 10/06/2010
你可以试试重叠 - 但它是无证功能, http://oraclesponge.wordpress.com/2008/06/12/the-overlaps-谓词/及其唯一检查数据范围是否重叠 – 2010-08-13 12:31:53
您能否显示表中列值的实际列/行示例以及此SQL查询返回的数据值 – 2010-08-13 12:33:03
结果中的日期交集是哪两个名字? A和C是不同的名称,B的两个范围似乎没有问题的日期范围。 – 2010-08-13 12:33:12