在SQL中查找日期范围重叠的记录 [英] Find records with overlapping date range in SQL

问题描述

我有以下表格和数据:

CREATE TABLE customer_wer(
  id_customer NUMBER,
  name VARCHAR2(10),
  surname VARCHAR2(20),
  date_from DATE,
  date_to DATE NOT NULL,
  CONSTRAINT customer_wer_pk PRIMARY KEY (id_customer, data_from));

INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '01-JAN-00', '31-MAR-00');
INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '01-APR-00', '30-JUN-00');
INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '15-JUN-00', '30-SEP-00');
INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '01-OCT-00', '31-DEC-00');
INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '01-JAN-01', '31-MAR-01');
INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '01-APR-01', '30-JUN-01');
INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '01-JUL-01', '5-OCT-01');
INSERT INTO customer_wer VALUES (4, 'Karolina', 'Komuda', '01-OCT-01', '31-DEC-01');

我需要一个SELECT查询来查找具有重叠日期的记录.这意味着在上面的示例中，我应该在结果中包含四个记录

I need a SELECT query to find the records with overlapping dates. It means that in the example above, I should have four records in result

number 
2
3
7
8

先谢谢您. 我正在使用Oracle DB.

Thank you in advance. I am using Oracle DB.

推荐答案

尝试一下:

select * from t t1
join t t2 on (t1.datefrom > t2.datefrom and t1.datefrom < t2.dateto)
          or (t1.dateto > t2.datefrom and t1.dateto < t2.dateto)

感谢您的示例.修改后，它可以正常工作:

Thank You for this example. After modification it is working:

SELECT *
FROM customer_wer k
JOIN customer_wer w
ON k.id_customer = w.id_customer
WHERE (k.date_from > w.date_to AND k.date_from < w.date_to)
OR (k.date_to > w.date_from AND k.date_to < w.date_to);

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