查找日期间隔

Question

下面是我的 table 数据：

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');
INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-JUL-01', '30-SEP-01');
INSERT INTO customer_wer VALUES (2, 'Karolina', 'Bielawska', '01-OCT-01', '31-DEC-01');

--LUKASZ JAGIELLO - GAP JUN JUL 2000

INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-JAN-00', '31-MAR-00');
INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-APR-00', '15-JUN-00');
INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-JUL-00', '30-SEP-00');
INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-OCT-00', '31-DEC-00');
INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-JAN-01', '31-MAR-01');
INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-APR-01', '30-JUN-01');
INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-JUL-01', '30-SEP-01');
INSERT INTO customer_wer VALUES (3, 'Lukasz', 'Jagiello', '01-OCT-01', '31-DEC-01');

我发现与以下 select:

的差距

SELECT date_from, date_to, id_customer, next_date
  FROM 
    (SELECT k.*,
          LEAD(date_from, 1) OVER (partition by id_customer order by date_from) AS next_date
      FROM customer_wer k
    )
WHERE date_to + 1 <> next_date AND date_to < '31-DEC-01' AND date_to < next_date;

我想知道复制 table 是否可以得到相同的结果。我找不到正确的解决方案。我的select如下：

SELECT COUNT(k.id_customer)
  FROM customer_wer k
  JOIN customer_wer w
  ON k.id_customer = w.id_customer
  WHERE (w.date_from < k.date_from AND w.date_to + 1 < k.date_from);

应该还有一个限制。

Answer 1

这个自连接查询产生了与您相同的输出：

with data as (
  select rownum rn, x.* from customer_wer x 
    order by id_customer, date_from)
select k.date_from, k.date_to, k.id_customer, w.date_from next_date
  from data k 
  join data w on k.id_customer = w.id_customer and k.rn + 1 = w.rn
  where k.date_to+1 < w.date_from
  order by k.id_customer, k.date_from;

而且我认为您应该将分析函数中的顺序更改为：

OVER (partition by id_customer order by date_from)

查找日期间隔

Finding gaps in dates

sql

oracle

date-arithmetic

oracle11g