2008-10-24 108 views
1

我有一个志愿者_2009表,列出了所有的志愿者和场地表,列出了志愿者可以分配到的场所,他们只被分配到一个。如何根据其他表格计算每个值的计数?

我想要做的是打印出分配给每个场地的志愿者人数。

我希望它打印出来是这样的:

地点的名称:志愿者#

表:volunteers_2009 列:ID,姓名,venue_id

表:场馆 列:ID ,venue_name

它们涉及通过volunteers_2009.venue_id = venues.id

这是我的,但它不正常工作。

$sql = "SELECT venues.venue_name as 'Venue', COUNT(volunteers_2009.id) as 'Number Of 
Volunteers' FROM venues ven JOIN volunteers_2009 vol ON 
(venues.id=volunteers_2009.venue_id) GROUP BY venues.venue_name ORDER BY 
venues.venue_name ASC"; 

$result = mysql_query($sql); 

while(list($name,$vols) = mysql_fetch_array($result)) { 
    print '<p>'.$name.': '.$vols.'</p>'; 
} 

回答

2

不是MySQL的人所以这可能是真的错了,但是当你给你的表的别名,你不那么需要用这个名字来引用它。

$sql = "SELECT ven.venue_name as 'Venue', COUNT(vol.id) as 'Number Of 
Volunteers' FROM venues ven JOIN volunteers_2009 vol ON 
(ven.id=vol.venue_id) GROUP BY ven.venue_name ORDER BY ven.venue_name ASC"; 
0
$query = "SELECT ven.venue_name AS 'Venue', count(*) AS 'Number of venues' 
     FROM volunteers_2009 AS vol, venues AS ven WHERE vol.venue_id = ven.id 
     GROUP BY ven.venue_name";