您可以在一个大喊大叫的声明中做到这一点:
SQL> WITH timeline AS
2 (SELECT mydate startdate,
3 lead(mydate) OVER (ORDER BY mydate) - 1 enddate
4 FROM (SELECT startdate mydate FROM interval_test
5 UNION
6 SELECT enddate FROM interval_test)
7 WHERE mydate IS NOT NULL)
8 SELECT startdate,
9 enddate,
10 max(substr(sys_connect_by_path(item, ','), 2)) items
11 FROM (SELECT t.startdate,
12 t.enddate,
13 item,
14 row_number() OVER (PARTITION BY t.startdate, t.enddate
15 ORDER BY i.item) rn
16 FROM timeline t
17 JOIN
18 interval_test i
19 ON nvl(i.enddate, DATE '9999-12-31') - 1 >= t.startdate
20 AND i.startdate <= nvl(t.enddate, DATE '9999-12-31'))
21 START WITH rn = 1
22 CONNECT BY rn = PRIOR rn + 1
23 AND startdate = PRIOR startdate
24 GROUP BY startdate, enddate
25 ORDER BY startdate;
STARTDATE ENDDATE ITEMS
---------- ---------- --------------------
2012-01-01 2012-01-31 AAA
2012-02-01 2012-02-29 AAA,BBB
2012-03-01 AAA
我使用第一个子查询列出所有间隔:
SQL> SELECT mydate startdate,
2 lead(mydate) OVER (ORDER BY mydate) - 1 enddate
3 FROM (SELECT startdate mydate FROM interval_test
4 UNION
5 SELECT enddate FROM interval_test)
6 WHERE mydate IS NOT NULL;
STARTDATE ENDDATE
---------- ----------
2012-01-01 2012-01-31
2012-02-01 2012-02-29
2012-03-01
加入以下查询,该查询列出了给定两个日期的一行中的所有项目:
SELECT max(substr(sys_connect_by_path(item, ','), 2)) items
FROM (SELECT item, row_number() OVER (ORDER BY item) rn
FROM interval_test
WHERE nvl(enddate, DATE '9999-12-31') >= :startdate
AND startdate <= :enddate)
CONNECT BY rn = PRIOR rn + 1
START WITH rn = 1;