通过 postgresql 循环获取活跃行程的数量

Question

我正在寻找一种方法，通过 postgresql 查询获取 GTFS 提要中一天中每一分钟的活跃行程量。

对于每一次旅行，我都有非规范化的开始和结束时间（以秒为单位）table。它看起来像这样：

为我提供在给定时间范围内有效的行程的查询（例如，此处为 43000 到 43600 秒）如下所示：

SELECT
COUNT(trips.trip_id)
FROM denormalized_trips AS trips

LEFT JOIN gtfs_calendar_dates AS calendar_dates
ON calendar_dates.service_id = trips.service_id
AND calendar_dates.agency_key = trips.agency_key
AND date = '2017-07-03'
AND exception_type = 1

INNER JOIN gtfs_calendar AS calendar
  ON trips.service_id = calendar.service_id
  AND calendar.agency_key = trips.agency_key
  AND calendar.wednesday = 1

WHERE (
  trip_start_time BETWEEN 46800 AND 47100
  AND '2017-07-03' BETWEEN calendar.start_date AND calendar.end_date
)
AND NOT EXISTS (
  SELECT 0
  FROM gtfs_calendar_dates AS date_exceptions
  WHERE date = '2017-07-03'
  AND date_exceptions.agency_key = trips.agency_key
  AND date_exceptions.service_id = calendar.service_id
  AND exception_type = 2
);

这将导致在 13:00pm 和 13:05pm 之间开始的 12 次行程。

现在我想一整天都这样做。我想获得在例如1 分钟或者更像是 5 分钟。我用一个循环试了一下，但这似乎只给了我 1 个结果。这是我现在想到的：

CREATE OR REPLACE FUNCTION get_active_trips(n int)
RETURNS INTEGER AS
$BODY$
DECLARE 
 count INTEGER;
BEGIN 
  FOR counter IN 43130..50000 BY 60 LOOP
    SELECT
    COUNT(trips.trip_id)
    INTO count
    FROM denormalized_trips AS trips

    LEFT JOIN gtfs_calendar_dates AS calendar_dates
    ON calendar_dates.service_id = trips.service_id
    AND calendar_dates.agency_key = trips.agency_key
    AND date = '2017-07-03'
    AND exception_type = 1

    INNER JOIN gtfs_calendar AS calendar
      ON trips.service_id = calendar.service_id
      AND calendar.agency_key = trips.agency_key
      AND calendar.wednesday = 1

    WHERE (
      trip_start_time BETWEEN counter AND counter + 60
      AND '2017-07-03' BETWEEN calendar.start_date AND calendar.end_date
    )
    AND NOT EXISTS (
      SELECT 0
      FROM gtfs_calendar_dates AS date_exceptions
      WHERE date = '2017-07-03'
      AND date_exceptions.agency_key = trips.agency_key
      AND date_exceptions.service_id = calendar.service_id
      AND exception_type = 2
    );
  END LOOP;
  RETURN count;
END; 
$BODY$ LANGUAGE plpgsql STABLE;

调用结果SELECT get_active_trips(1);

现在我想返回类似 table 的结果或一组结果，而不仅仅是 1 个条目。我该怎么做？

非常感谢任何帮助。

Answer 1

创建如下类型

drop type if exists get_active_trips_out;
create type get_active_trips_out as
(
   Count int
);

在从如下函数返回数据时使用 table 类型

CREATE OR REPLACE FUNCTION get_active_trips(n int)
RETURNS setof get_active_trips_out AS
$BODY$
DECLARE 
 count INTEGER;
r get_active_trips_out%rowtype;
BEGIN 
 create temp table tmp_count
(
  Count int
)on commit drop;
  FOR counter IN 43130..50000 BY 60 LOOP
    Insert into tmp_count
    SELECT
    COUNT(trips.trip_id)
    FROM denormalized_trips AS trips

    LEFT JOIN gtfs_calendar_dates AS calendar_dates
    ON calendar_dates.service_id = trips.service_id
    AND calendar_dates.agency_key = trips.agency_key
    AND date = '2017-07-03'
    AND exception_type = 1

    INNER JOIN gtfs_calendar AS calendar
      ON trips.service_id = calendar.service_id
      AND calendar.agency_key = trips.agency_key
      AND calendar.wednesday = 1

    WHERE (
      trip_start_time BETWEEN counter AND counter + 60
      AND '2017-07-03' BETWEEN calendar.start_date AND calendar.end_date
    )
    AND NOT EXISTS (
      SELECT 0
      FROM gtfs_calendar_dates AS date_exceptions
      WHERE date = '2017-07-03'
      AND date_exceptions.agency_key = trips.agency_key
      AND date_exceptions.service_id = calendar.service_id
      AND exception_type = 2
    );
  END LOOP;
 for r in 
 Select * from tmp_count
 loop
 return next r
 end loop;
END;

Answer 2

创建函数 returning 值集有两种语法：returns setof <type> 和 returns table(<columns definition>)。 Documentation.

还有几种方法可以从 plpgsql 函数中 return 这些值：return next 或 return query。 Documentation.

因此，如果您只想获得整数系列，您可以用这种方式重写您的函数：

CREATE OR REPLACE FUNCTION get_active_trips(n int)
RETURNS SETOF INTEGER AS
$BODY$
DECLARE 
 count INTEGER;
BEGIN 
  FOR counter IN 43130..50000 BY 60 LOOP
    SELECT
    COUNT(trips.trip_id)
    INTO count
    FROM denormalized_trips AS trips
    <rest of query here>
    ;
    RETURN NEXT count;
  END LOOP;
  RETURN;
END; 
$BODY$ LANGUAGE plpgsql STABLE;

或者，使用 RETURN QUERY:

CREATE OR REPLACE FUNCTION get_active_trips(n int)
RETURNS SETOF INTEGER AS
$BODY$
BEGIN 
  FOR counter IN 43130..50000 BY 60 LOOP
    RETURN QUERY
      SELECT
      COUNT(trips.trip_id)
      FROM denormalized_trips AS trips
      <rest of query here>
      ;
  END LOOP;
  RETURN;
END; 
$BODY$ LANGUAGE plpgsql STABLE;

如果你想 return 多一列：

CREATE OR REPLACE FUNCTION get_active_trips(n int)
RETURNS TABLE (counter_value int, active_trips_count int) AS
$BODY$
BEGIN 
  FOR counter IN 43130..50000 BY 60 LOOP
    SELECT
    COUNT(trips.trip_id)
    INTO active_trips_count
    FROM denormalized_trips AS trips
    <rest of query here>
    ;
    counter_value := counter;
    RETURN NEXT; -- There is no parameters, current values of counter_value and active_trips_count will be returned
  END LOOP;
  RETURN;
END; 
$BODY$ LANGUAGE plpgsql STABLE;

或者，使用 RETURN QUERY:

CREATE OR REPLACE FUNCTION get_active_trips(n int)
RETURNS TABLE (counter_value int, active_trips_count int) AS
$BODY$
BEGIN 
  FOR counter IN 43130..50000 BY 60 LOOP
    RETURN QUERY
      SELECT
        counter,
        COUNT(trips.trip_id)
      FROM denormalized_trips AS trips
      <rest of query here>
      ;
  END LOOP;
  RETURN;
END; 
$BODY$ LANGUAGE plpgsql STABLE;

终于有了 returns table 的替代声明：

CREATE OR REPLACE FUNCTION get_active_trips(
  n int,
  out counter_value int, 
  out active_trips_count int)
RETURNS SETOF RECORD AS

更新

但是(!)我觉得可以使用没有循环的单个查询来简化您的任务。

考虑以下查询（我使用了 sqlfiddle 中的简化查询）：

select
  counter,
  count(trips.trip_id)
from
  generate_series(43130, 50000, 60) as counter left join
    denormalized_trips as trips on (trip_start_time between counter and counter + 60) 
group by counter 
order by counter;

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