Question

p_id    book_num conf_num   arrival_dt  departure_dt    create-dt   room_num
353     21807    3328568    19-JUN-15   21-JUN-15       27-JUN-15   2408
353     21807    3328562    18-JUN-15   20-JUN-15       27-JUN-15   2408

在上面的示例中，arrival_dt和departure_dt重叠为2个不同同一房间号码2408的确认号码

我还要排除以下记录集，其中arrival_dt和departure_dt相同

p_id    book_num conf_num   arrival_dt  departure_dt    create-dt   room_num
353     21802    3328508    18-JUN-15   21-JUN-15       27-JUN-15   1909    
353     21802    3328555    18-JUN-15   21-JUN-15       27-JUN-15   1909

请你帮我用SQL逻辑在表格中找到这些记录

Answer 1

正确的逻辑是，一个人在另一个人到达之后离开而第一个人在另一个人离开之前到达。您可以使用自联接或where子句来执行此操作。

如果您只想要记录：

select r.*
from records r
where exists (select 1
              from records r2
              where r2.pid = r.pid and
                    r2.arrival_dt >= r.departure_dt and
                    r2.departure_dt <= r.arrival_dt
             );

Answer 2

SELECT tbl.*
FROM table tbl
JOIN table tbl1 ON tbl.p_id= tbl1.p_id
WHERE tbl.Arrival_dt <= tbl1.Departure_dt
    AND tbl.Departure_dt >= tbl1.Arrival_dt

Answer 3

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE TEST ( p_id, book_num, conf_num, arrival_dt, departure_dt, create_dt, room_num ) AS
          SELECT 353, 21807, 3328568, DATE '2015-06-19', DATE '2015-06-21', DATE '2015-06-27', 2408 FROM DUAL
UNION ALL SELECT 353, 21807, 3328562, DATE '2015-06-18', DATE '2015-06-20', DATE '2015-06-27', 2408 FROM DUAL
UNION ALL SELECT 353, 21802, 3328508, DATE '2015-06-18', DATE '2015-06-21', DATE '2015-06-27', 1909 FROM DUAL
UNION ALL SELECT 353, 21802, 3328555, DATE '2015-06-18', DATE '2015-06-21', DATE '2015-06-27', 1909 FROM DUAL
UNION ALL SELECT 353, 21801, 3328444, DATE '2015-06-17', DATE '2015-06-21', DATE '2015-06-27', 2000 FROM DUAL
UNION ALL SELECT 353, 21801, 3328445, DATE '2015-06-18', DATE '2015-06-20', DATE '2015-06-27', 2000 FROM DUAL
UNION ALL SELECT 353, 21803, 3328446, DATE '2015-06-19', DATE '2015-06-20', DATE '2015-06-27', 2001 FROM DUAL
UNION ALL SELECT 353, 21804, 3328447, DATE '2015-06-20', DATE '2015-06-21', DATE '2015-06-27', 2001 FROM DUAL;

查询1 ：

SELECT *
FROM   TEST t
WHERE  EXISTS ( SELECT 'X'
                FROM   TEST x
                WHERE  x.room_num     = t.room_num
                AND    x.arrival_dt   < t.departure_dt
                AND    x.departure_dt > t.arrival_dt
                AND NOT (    x.arrival_dt   = t.arrival_dt
                         AND x.departure_dt = t.departure_dt ) )

<强> Results ：

| P_ID | BOOK_NUM | CONF_NUM |             ARRIVAL_DT |           DEPARTURE_DT |              CREATE_DT | ROOM_NUM |
|------|----------|----------|------------------------|------------------------|------------------------|----------|
|  353 |    21807 |  3328568 | June, 19 2015 00:00:00 | June, 21 2015 00:00:00 | June, 27 2015 00:00:00 |     2408 |
|  353 |    21807 |  3328562 | June, 18 2015 00:00:00 | June, 20 2015 00:00:00 | June, 27 2015 00:00:00 |     2408 |
|  353 |    21801 |  3328444 | June, 17 2015 00:00:00 | June, 21 2015 00:00:00 | June, 27 2015 00:00:00 |     2000 |
|  353 |    21801 |  3328445 | June, 18 2015 00:00:00 | June, 20 2015 00:00:00 | June, 27 2015 00:00:00 |     2000 |

从数据集中查找重叠的日期范围

3 个答案: