用CTE對查詢分組結果

我有一個基於CTE的查詢，其中我傳遞了大約2600個4元組緯度/經度值 - 這些值已被標記並保存在稱爲座標的第二個表中。將這些左上角和右下角經度/緯度值傳遞給CTE，以顯示給定兩個時間戳的請求數量（小時）。用CTE對查詢分組結果

但是，我想在給定的時間戳內獲得每天的總請求數。也就是說，我想獲得每個指定日期的用戶請求總數。例如。用戶選擇每個星期三或星期三和星期四等 - 在2012年1月1日至16日之間的11:55和22:04之間，查看我經過的每個緯度/經度4元組。輸出基本上是這樣的：

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) 
. 
. 
.

我該怎麼做？我的查詢如下：

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;

和取得的樣品結果是這樣的：

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，我想看看這是

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

來源

2012-10-04 sm90901

將您的最終SELECT更改爲：

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;

迄今爲止鑄造cal.stamp的關鍵部分：cal.stamp::date。
那，和sum(q.zcount)。

來源

2012-10-04 23:26:20

我剛剛看到一篇文章，您稱PostgreSQL的「Good Guy Greg」。我同意那個。 – sm90901

@ sm90901：恩，謝謝。 :) –

剛剛注意到一個小問題 - 實際上我忘了首先問這個問題 - 在我的「初始」版本中 - 當我只需要總查詢數和其他任何類似lat/lons-時，我的查詢還打印了具有0計數的行感謝'COALESCE（q.zcount，0）'。我如何將它應用到這種情況，以獲得zcount 0的行？我將'COALESCE'應用於'sum（q.zcount）'，但沒有按照我的計劃工作。 – sm90901

用CTE對查詢分組結果

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