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[英]BigQuery - Create a table from results of a query that uses complex CTEs?
[英]Grouping the results of a query with CTEs
我有一个基于CTE的查询,在其中传递了大约2600个4元组的纬度/经度值-这些值已ID标记并保存在称为坐标的第二个表中。 这些左上和右下的纬度/经度值传递到CTE中,以显示在给定的两个时间戳下这些坐标内发出的请求数量(每小时)。
但是,我想在给定的时间戳内每天获取总请求数。 也就是说,我想获取每个指定日期的用户请求总数。 例如,用户选择在我经过的每个纬度/经度4元组中查看每个星期三或星期三以及星期四等-在2012年1月1日至16日之间的11:55到22:04之间。 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,如上所述,我希望看到
coordinates_id | stamp | zcount
1 2012-01-04 2
1 2012-01-11 3
2 2012-01-04 1
将最终的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;
将cal.stamp
为日期的关键部分: cal.stamp::date
。
那和sum(q.zcount)
。
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