使用 PL/pgSQL 在循环中运行 CTE 查询

Question

我正在尝试执行使用 plpgsql 在循环中重复调用的查询 - 循环遍历另一个表（命名坐标），其中包含网格的左上角和右下角纬度/经度坐标，我传递左上角和右下角纬度/ 经度值输入我的 CTE，以便显示在给定两个时间戳的这些坐标内发出的请求量（每小时）-。但是，我无法显示 CTE 的结果，并且收到以下错误消息：

ERROR:  query has no destination for result data
HINT:  If you want to discard the results of a SELECT, use PERFORM instead.
CONTEXT:  PL/pgSQL function "inline_code_block" line 6 at SQL statement

为了使整个查询根据需要工作，我应该在这里更改什么？我的代码如下：

DO $$
<<outer_scope>> DECLARE
  coords RECORD;
BEGIN
    FOR coords IN SELECT topleftlat, topleftlon, bottomrightlat, bottomrightlon FROM coordinates LOOP
        WITH cal AS (
        SELECT generate_series('2011-02-02 00:00:00'::timestamp ,
                   '2012-04-01 05:00:00'::timestamp , 
                   '1 hour'::interval) AS stamp
    ),
    qqq AS (
      SELECT date_trunc('hour', calltime) AS stamp, count(*) AS zcount
      FROM mytable
      WHERE calltime >= '2011-02-13 11:55:11' 
        AND calltime <= '2012-02-13 01:02:21'
        AND (calltime::time >= '11:55:11' 
        OR calltime::time <= '01:02:21')
        AND lat BETWEEN coords.bottomrightlat AND coords.topleftlat
        AND lon BETWEEN coords.topleftlon AND coords.bottomrightlon
     GROUP BY date_trunc('hour', calltime)
    )
    SELECT cal.stamp, COALESCE (qqq.zcount, 0) AS zcount
    FROM cal
    LEFT JOIN qqq ON cal.stamp = qqq.stamp
    WHERE cal.stamp >= '2011-02-13 11:00:00' 
      AND cal.stamp <= '2012-02-13 01:02:21' 
      AND (
        extract ('hour' from cal.stamp) >= extract ('hour' from '2011-02-13 11:00:00'::timestamp) or
        extract ('hour' from cal.stamp) <= extract ('hour' from '2012-02-13 01:02:21'::timestamp) 
      )
    ORDER BY stamp ASC;

    END LOOP;
END;
$$;

Answer 1

该DO命令无法实际返回数据（除了使用RAISE，或者您可以写入（临时）表 .. ）。

您需要创建一个可以定义返回类型RETURNS并调用它的 PL/pgSQL 函数。

你可以用 返回结果RETURN QUERY EXECUTE。但我怀疑整个操作可以简化......

重写为单个 SQL 查询

您可能根本不需要 plpgsql 或循环。考虑这个简单的 SQL 查询：

WITH v AS (
   SELECT '2011-02-13 11:55:11'::timestamp AS _from -- provide times once
         ,'2012-02-13 01:02:21'::timestamp AS _to
   )
, q AS (
   SELECT c.coordinates_id
        , date_trunc('hour', t.calltime) AS stamp
        , count(*) AS zcount
   FROM   v
   JOIN   mytable t ON  t.calltime BETWEEN v._from AND v._to
                   AND (t.calltime::time >= v._from::time OR
                        t.calltime::time <= v._to::time)
   JOIN   coordinates c ON (t.lat, t.lon) 
                   BETWEEN (c.bottomrightlat, c.topleftlon)
                       AND (c.topleftlat, c.bottomrightlon)
   GROUP  BY c.coordinates_id, date_trunc('hour', t.calltime)
   )
, cal AS (
   SELECT generate_series(GREATEST('2011-02-02 00:00:00'::timestamp, v._from)
                        , LEAST('2012-04-01 05:00:00'::timestamp, v._to)
                        , '1 hour'::interval) AS stamp
   FROM v
   )
SELECT q.coordinates_id, cal.stamp, COALESCE (q.zcount, 0) AS zcount
FROM   v, cal
LEFT   JOIN q USING (stamp)
WHERE (cal.stamp::time >= v._from::time OR
       cal.stamp::time <= v._to::time)
ORDER  BY q.coordinates_id, stamp;

• 不要循环遍历 table 中的行，而是coordinates加入 CTE 并一次性生成整个结果。

• 当您聚合每一行时，coordinates我们需要这个表的主键（或任何其他唯一的列集），我假设一个名为coordinates_id.

• 我加入了CTE v（即“值”）之上，以提供_from与_to仅一次时间戳。

• 我立即使用_from和_to来限制日历的时间范围，而不是WHERE在最终的SELECT.

  GREATEST('2011-02-02 00:00:00'::timestamp, v._from)
  LEAST('2012-04-01 05:00:00'::timestamp, v._to)
  

• 对于更简单的条件，我使用@kgrittn 在此相关答案中演示的“临时行” JOIN：

       ON (t.lat, t.lon) 
  BETWEEN (c.bottomrightlat, c.topleftlon)
      AND (c.topleftlat, c.bottomrightlon)
  

• 我转换为 time ( ::time) 而不是使用extract ('hour' ..)，因为它更简单、更快。

我不是 100% 确定这正是您所追求的，但它应该非常接近。