如何在 Postgres 中查找连续的日期 SQL

Question

我在 postgres 数据库中有以下 table（table 名称是 table_test）：

          id             dia          Data_sensor_Analog
         2165         2020-09-20       4585542
         2165         2020-09-21       4954566
         2165         2020-09-26           255

我想统计连续多少天有属性dia。

为此，我尝试编写了以下代码：

           WITH 

           groups AS (
           SELECT
              ROW_NUMBER() OVER (ORDER BY dia) AS rn,
              dateadd(dia, -ROW_NUMBER() OVER (ORDER BY dia), dia) AS grp,
              dia
           FROM table_test
          )

          SELECT
          COUNT(*) AS consecutiveDates,
          MIN(dia) AS minDate,
          MAX(dia) AS maxDate
          FROM groups
          GROUP BY grp
          ORDER BY 1 DESC, 2 DESC

我希望输出为：

             consecutiveDates       minDate        maxDate  
                     1            2020-09-20      2020-09-21

但是，当我运行代码时，出现如下错误信息：

          ERROR:  function dateadd(text, bigint, text) does not exist
          LINE 17:       dateadd(dia, -ROW_NUMBER() OVER (ORDER BY dia), dia) A

我正在使用 postgres，并在网站上找到了这个示例代码：https://blog.jooq.org/2015/11/07/how-to-find-the-longest-consecutive-series-of-events-in-sql/

我将 dia 属性转换为：

         ALTER TABLE table_test
         ALTER COLUMN dia
         TYPE TIMESTAMP WITHOUT TIME ZONE
         USING dia::timestamp without time zone;

Answer 1

你可以减去一个枚举值，但你需要一个子查询或CTE：

select min(dia), max(dia), count(*)
from (select t.*,
             row_number() over (order by dia) as seqnum
      from table_test t
     ) t
group by dia - seqnum * interval '1 day';

但是，看起来 dia 是一个字符串而不是日期。要解决这个问题：

group by (dia::date) - seqnum * interval '1 day';

格式适合转换为日期。

Here 是一个 db<>fiddle.

Answer 2

考虑到您的 table 中一天只有一个条目，那么试试这个：

select id, count(*) -1 "count", max(dia), min(dia) from (
select *, 
date(dia) - row_number() over (partition by id order by date(dia)) * interval '1 day' "filter" 
from table_test
) t1 
group by id, filter
having count(*) -1 > 0

DEMO

如果同一日期有多个值，请尝试以下操作：

with cte as (
select 
*,
date(dia) date_,date(dia) - dense_rank() over ( partition by id order by date(dia)) * interval '1 day' "filter" 
from table_test
)
select 
id, count(distinct date_) -1 "count" , max(dia),min(dia) 
from cte
group by id, filter
having count(distinct date_) -1 >0

DEMO

如何在 Postgres 中查找连续的日期 SQL

How to Find Consecutive Dates in Postgres SQL

sql

postgresql

window-functions