小编Luk*_*asz的帖子

找出日期的差距

下面是我的数据表:

CREATE TABLE customer_wer (
  id_customer NUMBER,
  name VARCHAR2(10),
  surname VARCHAR2(20),
  date_from DATE,
  date_to DATE);

--KAROLINA BIELAWSKA - GAP MAR APR 2000 and JUN JUL 2001

INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-JAN-00', '28-MAR-00');
INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-APR-00', '30-JUN-00');
INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-JUL-00', '30-SEP-00');
INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-OCT-00', '31-DEC-00');
INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-JAN-01', '31-MAR-01');
INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-APR-01', '15-JUN-01'); …

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sql oracle date-arithmetic oracle11g

Luk*_*asz

2015 04-30

5
推荐指数

1
解决办法

73
查看次数

在 SQL 中查找日期范围重叠的记录

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 …

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sql oracle

Luk*_*asz

2015 05-02

3
推荐指数

1
解决办法

2万
查看次数