MySQL的：使用GROUP BY時

我有MySQL表MySQL的：使用GROUP BY時

CREATE TABLE cms_webstat (
    ID int NOT NULL auto_increment PRIMARY KEY, 
    TIMESTAMP_X timestamp DEFAULT CURRENT_TIMESTAMP, 
    # ... some other fields ... 
)

其中包含有關網站訪問者的統計數據填充空字段用零。
爲了得到每小時訪問我用

SELECT 
    hour(TIMESTAMP_X) as HOUR 
    , count(*) AS HOUR_STAT 
FROM cms_webstat 
GROUP BY HOUR 
ORDER BY HOUR DESC

這給了我

| HOUR | HOUR_STAT | 
| 24 | 15  | 
| 23 | 12  | 
| 22 | 9  | 
| 20 | 3  | 
| 18 | 2  | 
| 15 | 1  | 
| 12 | 3  | 
| 9 | 1  | 
| 3 | 5  | 
| 2 | 7  | 
| 1 | 9  | 
| 0 | 12  |

而且我想獲得如下：

| HOUR | HOUR_STAT | 
| 24 | 15  | 
| 23 | 12  | 
| 22 | 9  | 
| 21 | 0  | 
| 20 | 3  | 
| 19 | 0  | 
| 18 | 2  | 
| 17 | 0  | 
| 16 | 0  | 
| 15 | 1  | 
| 14 | 0  | 
| 13 | 0  | 
| 12 | 3  | 
| 11 | 0  | 
| 10 | 0  | 
| 9 | 1  | 
| 8 | 0  | 
| 7 | 0  | 
| 6 | 0  | 
| 5 | 0  | 
| 4 | 0  | 
| 3 | 5  | 
| 2 | 7  | 
| 1 | 9  | 
| 0 | 12  |

我應該如何修改查詢以獲得這樣的結果（與一個MySQL查詢，不創建臨時表）？
是否有可能通過一個MySQL查詢獲得這樣的結果？

來源

2010-03-19 SaltLake

SaltLake，你認爲你會接受答案嗎？ – 2010-03-22 16:30:16

馬庫斯，你的解決方案的工作，但不是我想要的方式。我對使用一個SQL查詢解決方案感興趣，並且不需要創建和填充另一個表。 – SaltLake 2010-03-23 00:39:20

我已經找到了答案。也許我瘋了，但這個工程。

 
SELECT HOUR, max(HOUR_STAT) as HOUR_STAT FROM (
    (
     SELECT HOUR(TIMESTAMP_X) as HOUR, count(*) as HOUR_STAT 
     FROM cms_webstat 
     WHERE date(TIMESTAMP_X) = date(now()) 
    ) 
    UNION (SELECT 0 as HOUR, 0) 
    UNION (SELECT 1 as HOUR, 0) 
    UNION (SELECT 2 as HOUR, 0) 
    UNION (SELECT 3 as HOUR, 0) 
    UNION (SELECT 4 as HOUR, 0) 
    UNION (SELECT 5 as HOUR, 0) 
    UNION (SELECT 6 as HOUR, 0) 
    UNION (SELECT 7 as HOUR, 0) 
    UNION (SELECT 8 as HOUR, 0) 
    UNION (SELECT 9 as HOUR, 0) 
    UNION (SELECT 10 as HOUR, 0) 
    UNION (SELECT 11 as HOUR, 0) 
    UNION (SELECT 12 as HOUR, 0) 
    UNION (SELECT 13 as HOUR, 0) 
    UNION (SELECT 14 as HOUR, 0) 
    UNION (SELECT 15 as HOUR, 0) 
    UNION (SELECT 16 as HOUR, 0) 
    UNION (SELECT 17 as HOUR, 0) 
    UNION (SELECT 18 as HOUR, 0) 
    UNION (SELECT 19 as HOUR, 0) 
    UNION (SELECT 20 as HOUR, 0) 
    UNION (SELECT 21 as HOUR, 0) 
    UNION (SELECT 22 as HOUR, 0) 
    UNION (SELECT 23 as HOUR, 0) 
) 
AS `combined_table` 
GROUP BY HOUR 
ORDER BY HOUR DESC

根據需要的一個MySQL查詢。

來源

2010-10-13 09:39:55 SaltLake

創建一個列另一個表，

CREATE TABLE hours_list (
    hour int NOT NULL PRIMARY KEY 
)

所有24小時填充它。

然後在該表上進行連接以填充零。

SELECT 
    hs.hour as HOUR, COUNT(ws.ID) AS HOUR_STAT 
FROM hours_list hs 
LEFT JOIN cms_webstat ws ON hs.hour = hour(ws.TIMESTAMP_X) 
GROUP BY hs.hour 
ORDER BY hs.hour DESC

來源

2010-03-19 18:46:22

這只是'爲什麼它沒有返回'部分。馬庫斯的答案涵蓋了'如何'部分。

的SQL

SELECT 
    hour(TIMESTAMP_X) as HOUR 
    , count(*) AS HOUR_STAT 
FROM cms_webstat 
GROUP BY HOUR 
ORDER BY HOUR DESC

得到的記錄每小時計，爲表中存在

它不給，什麼是不存在於表中的細節的時間戳。由於沒有記錄與時間8（來自您的示例）相對應的時間戳記，SQL不會返回任何記錄。

來源

2010-10-13 08:57:02 Nivas

$sql = 'SELECT g, MAX(v) AS v, MAX(c) AS c FROM ('; 
$sql .= '(SELECT DATE_FORMAT(viewed, \'%d.%m.%Y\') AS g, COUNT(1) AS v, 0 AS c FROM '.$this->prefix.'view WHERE campaignid IN ('.join(', ',$ids).') GROUP BY g)'; 
$sql .= ' UNION (SELECT DATE_FORMAT(clicked, \'%d.%m.%Y\') AS g, 0 AS v, COUNT(1) AS c FROM '.$this->prefix.'clicks WHERE campaignid IN ('.join(', ',$ids).') GROUP BY g)'; 
$today = strtotime("00:00:00"); 
for ($i=$today; $i>=time()-30*86400; $i-=86400) { 
    $sql .= ' UNION (SELECT \''.date('d.m.Y',$i).'\' AS g, 0 AS v, 0 AS c)'; 
} 
$sql .= ') AS tmp GROUP BY g ORDER BY g DESC'; 

$chart = DB::getAll($sql); 
p($chart);

謝謝！做好了！從2個表格，點擊和視圖，加入..作品。 ajaxel.com

來源

2014-05-30 19:35:02 Alexander

MySQL的：使用GROUP BY時

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