我怎样才能做出这样的事情(在 SELECT 命令中添加一些东西):
 $query = "SELECT a.name,a.surname,b.email,c.phone 
           FROM users as a 
           inner join users_email as b 
           inner join users_phone as c
           WHERE a.id=b.id AND a.id=c.id";
 $query .= "ORDER BY a.surname";  
 $result = mysql_query($query,$con);
结果应该是:
$query = "SELECT a.name,a.surname,b.email,c.phone 
          FROM users as a 
          inner join users_email as b 
          inner join users_phone as c
          WHERE a.id=b.id AND a.id=c.id
          ORDER BY a.name";   
$result = mysql_query($query,$con);
$row=mysql_num_fields($result)
...但它给出了一个错误:mysql_num_fields() 期望参数 1 是资源,在第 32 行的 C:\xampp-portable\htdocs\db\file.php 中给出的布尔值...