连续日期postgresql

Question

我需要知道每个文档是否没有连续的日期。我有这个 table:

document | the_day  
     1   | 2015-01-01  
     1   | 2015-01-02  
     1   | 2015-01-03  
     1   | 2015-01-04  
     1   | 2015-01-05  
     1   | 2015-01-06  
     2   | 2015-01-01  
     2   | 2015-01-03  
     2   | 2015-01-04  
     2   | 2015-01-05  
     2   | 2015-01-06  
     3   | 2015-01-01  
     3   | 2015-01-02  
     3   | 2015-01-03  
     3   | 2015-01-04  
     3   | 2015-01-05  
     3   | 2015-01-06

正如您所看到的，只有一个差距：在文档 2 中缺少“2015-01-02”。我想知道这个差距。我有这个 select:

SELECT document, the_day, the_day - lag(the_day) OVER w AS gap
              FROM mytable
              where active=true and fault=false
               WINDOW w AS (ORDER BY document,the_day)

这个 select 给我一个每个日期的寄存器，在大多数情况下，差距是 1，但是当另一个文档开始出现在结果中时，它给我的差距是错误的。我不知道这是正确的方法还是使功能... 这里是构建 table:

的代码

--Table: public.test_consecutives

--DROP TABLE public.test_consecutives;

CREATE TABLE public.test_consecutives (
  document  integer,
  the_day   date
) WITH (
    OIDS = FALSE
  );

ALTER TABLE public.test_consecutives
  OWNER TO postgres;
INSERT INTO test_consecutives (document, the_day) VALUES
    (1, '2015-01-01'),
    (1, '2015-01-02'),
    (1, '2015-01-03'),
    (1, '2015-01-04'),
    (1, '2015-01-05'),
    (1, '2015-01-06'),
    (2, '2015-01-01'),
    (2, '2015-01-03'),
    (2, '2015-01-04'),
    (2, '2015-01-05'),
    (2, '2015-01-06'),
    (3, '2015-01-01'),
    (3, '2015-01-02'),
    (3, '2015-01-03'),
    (3, '2015-01-04'),
    (3, '2015-01-05'),
    (3, '2015-01-06');

Answer 1

如果您不指定 PARTITION，PostgreSQL 将假定它是整个 table。您的查询应包含 PARTITION BY 子句：

SELECT document, the_day, the_day - lag(the_day) OVER w AS gap
    FROM mytable
    where active=true and fault=false
    WINDOW w AS (PARTITION BY document ORDER BY document,the_day)

连续日期postgresql

Consecutive dates postgresql

sql

postgresql

gaps-and-islands