Question

我有一个基于CTE的查询，我在其中传递了大约2600个4元组纬度/经度值 - 这些值已被ID标记并保存在称为坐标的第二个表中。这些左上角和右下角的纬度/经度值将传递到CTE，以显示给定两个时间戳在这些坐标内进行的请求量（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，我希望在每个指定日期获得用户请求的总数。例如。用户选择查看每周三或周三和周四等 - 在2012年1月1日至16日期间的11:55和22:04之间，我通过的每个纬度/经度4元组。输出基本上就像：

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

所以，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

Answer 1

将您的最终SELECT更改为：

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;

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

使用CTE对查询结果进行分组

1 个答案: