ORACLE SQL 日期范围交集 [英] ORACLE SQL Date range intersections

问题描述

我有一个表 T1，它包含一个 NAME 值(不是唯一的)和一个日期范围(D1 和 D2 是日期)当 NAME 相同时，我们对日期范围进行联合(例如 B).

I have a table T1, it contains a NAME value (not unique), and a date range (D1 and D2 which are dates) When NAME are the same, we make a union of the date ranges (e.g. B).

但作为结果 (X)，我们需要对所有日期范围进行交集

But as a result (X), we need to make intersection of all the date ranges

表T1

NAME | D1       | D2
A    | 20100101 | 20101211
B    | 20100120 | 20100415
B    | 20100510 | 20101230
C    | 20100313 | 20100610

结果:

X    | 20100313 | 20100415
X    | 20100510 | 20100610

在视觉上，这将给出以下内容:

Visually, this will give the following :

NAME        : date range
A           : [-----------------------]-----
B           : --[----]----------------------
B           : ----------[---------------]---
C           : -----[--------]---------------

结果:

X           : -----[-]----------------------
X           : ----------[---]---------------

知道如何使用 SQL/PL SQL 获得它吗?

Any idea how to get that using SQL / PL SQL ?

推荐答案

这里有一个快速的解决方案(可能不是最有效的):

here is a quick solution (may not be the most efficient):

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

我构建了所有可能的连续日期范围，并将此日历"与示例数据结合起来.这将列出具有 3 个值的所有范围.如果添加行，您可能需要合并结果:

I build all the possible successive date ranges and join this "calendar" with the sample data. This will list all ranges that have 3 values. You may need to merge the result if you add rows:

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

这篇关于ORACLE SQL 日期范围交集的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！