MySQL的：使用GROUP BY时

我有MySQL表MySQL的：使用GROUP BY时

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

其中包含有关网站访问者的统计数据填充空字段用零。
为了得到每小时访问我用

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

这给了我

| 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  |

而且我想获得如下：

| 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  |

我应该如何修改查询以获得这样的结果（与一个MySQL查询，不创建临时表）？
是否有可能通过一个MySQL查询获得这样的结果？

来源

2010-03-19 SaltLake

SaltLake，你认为你会接受答案吗？ – 2010-03-22 16:30:16

马库斯，你的解决方案的工作，但不是我想要的方式。我对使用一个SQL查询解决方案感兴趣，并且不需要创建和填充另一个表。 – SaltLake 2010-03-23 00:39:20

我已经找到了答案。也许我疯了，但这个工程。

 
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

根据需要的一个MySQL查询。

来源

2010-10-13 09:39:55 SaltLake

创建一个列另一个表，

CREATE TABLE hours_list (
    hour int NOT NULL PRIMARY KEY 
)

所有24小时填充它。

然后在该表上进行连接以填充零。

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

来源

2010-03-19 18:46:22

这只是'为什么它没有返回'部分。马库斯的答案涵盖了'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

得到的记录每小时计，为表中存在

它不给，什么是不存在于表中的细节的时间戳。由于没有记录与时间8（来自您的示例）相对应的时间戳记，SQL不会返回任何记录。

来源

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);

谢谢！做好了！从2个表格，点击和视图，加入..作品。 ajaxel.com

来源

2014-05-30 19:35:02 Alexander

MySQL的：使用GROUP BY时

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