用CTE对查询分组结果

我有一个基于CTE的查询，其中我传递了大约2600个4元组纬度/经度值 - 这些值已被标记并保存在称为坐标的第二个表中。将这些左上角和右下角经度/纬度值传递给CTE，以显示给定两个时间戳的请求数量（小时）。用CTE对查询分组结果

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，我想获得每个指定日期的用户请求总数。例如。用户选择每个星期三或星期三和星期四等 - 在2012年1月1日至16日之间的11:55和22:04之间，查看我经过的每个纬度/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;

和取得的样品结果是这样的：

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，我想看看这是

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

来源

2012-10-04 sm90901

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;

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

来源

2012-10-04 23:26:20

我刚刚看到一篇文章，您称PostgreSQL的“Good Guy Greg”。我同意那个。 – sm90901

@ sm90901：恩，谢谢。 :) –

刚刚注意到一个小问题 - 实际上我忘了首先问这个问题 - 在我的“初始”版本中 - 当我只需要总查询数和其他任何类似lat/lons-时，我的查询还打印了具有0计数的行感谢'COALESCE（q.zcount，0）'。我如何将它应用到这种情况，以获得zcount 0的行？我将'COALESCE'应用于'sum（q.zcount）'，但没有按照我的计划工作。 – sm90901

用CTE对查询分组结果

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