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我有一个基于 CTE 的查询，我向其中传递了大约 2600 个 4 元组纬度/经度值 - 这些值已被 ID 标记并保存在名为坐标的第二个表中。这些左上角和右下角的纬度/经度值被传递到 CTE，以显示给定两个时间戳在这些坐标内发出的请求量（每小时）。

但是，我想在给定的时间戳内获得每天的总请求数。也就是说，我想获取每个指定日期的用户请求总数。例如，用户选择每周三或周三和周四等查看 - 2012 年 1 月 1 日至 16 日期间的 11:55 至 22:04 之间，我通过的每个纬度/经度 4 元组。输出基本上是这样的：

coordinates_id | stamp       | zcount

1                Jan 4 2012    200 (total requests on Wednesday Jan 4 between 11:55 and 22:04)
1                Jan 11 2012   121 (total requests on Wednesday Jan 11 between 11:55 and 22:04)
2                Jan 4 2012    255 (total requests on Wednesday Jan 4 between 11:55 and 22:04)
2                Jan 11 2012   211 (total requests on Wednesday Jan 11 between 11:55 and 22:04)
.
.
.

我该怎么做？我的查询如下：

WITH v AS (
   SELECT '2012-01-1 11:55:11'::timestamp AS _from -- provide times once
         ,'2012-01-16 22:02:21'::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 AND
                        t.calltime::time <= v._to::time) AND 
(extract(DOW from t.calltime) = 3)
   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('2011-2-2 00:00:00'::timestamp
                        , '2012-4-1 05:00:00'::timestamp
                        , '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 (extract(hour from cal.stamp) >= extract(hour from v._from) AND
       extract(hour from cal.stamp) <= extract(hour from v._to)) AND 
(extract(DOW from cal.stamp) = 3)
       AND cal.stamp >= v._from AND cal.stamp <= v._to
GROUP BY q.coordinates_id, cal.stamp, q.zcount
ORDER BY q.coordinates_id ASC, stamp ASC;

它产生的样本结果是这样的：

coordinates_id  | stamp                | zcount
1                 2012-01-04 16:00:00    1
1                 2012-01-04 19:00:00    1
1                 2012-01-11 14:00:00    1
1                 2012-01-11 17:00:00    1
1                 2012-01-11 19:00:00    1
2                 2012-01-04 16:00:00    1

因此，正如我上面提到的，我希望将其视为

coordinates_id  | stamp      | zcount
1                2012-01-04    2
1                2012-01-11    3
2                2012-01-04    1
4

1 回答 1

1

将您的最终更改SELECT为：

SELECT q.coordinates_id, cal.stamp::date, sum(q.zcount) AS zcount
FROM   v, cal
LEFT   JOIN q USING (stamp)
WHERE  extract(hour from cal.stamp) BETWEEN extract(hour from v._from)
                                        AND extract(hour from v._to)
AND    extract(DOW from cal.stamp) = 3 
AND    cal.stamp >= v._from
AND    cal.stamp <= v._to
GROUP  BY 1,2
ORDER  BY 1,2;

cal.stamp迄今为止，它的关键部分是： cal.stamp::date.
那，和sum(q.zcount)。

于 2012-10-04T23:26:20.817 回答