2013-05-26 31 views
0

我是php的初学者。和我有此SQL问题:php函数中的MySQL-Link错误

function InsertUserBirdsFromFile($File_content){ 
for($i =0; $i < count($File_content); $i+=2){ 
$id = $this->Master_file($File_content[$i], $File_content[$i + 1]); 
if(isset($id)){ 
try{ 
$qry = "insert into user_to_birds(user_id,tax_id)values(1 ,'.$id .') "; 
     $result = mysql_query($qry,$this->connection); 

} 
catch(Exception $ex){ echo $ex;} 
} 
} 
} 

function Master_file($name, $latin){ 
try{ 

$qry = "SELECT tax_id FROM master where name =".$name." and latin =".$latin; 
     $result = mysql_query($qry,$this->connection); 
     } 
     catch(Exception $ex){ return null;} 
         if ($result == true && mysql_num_rows($result) >0) { 
         $p=0; 
          while ($Res_user = mysql_fetch_array($result)) { 
         $marques[$p] = $Res_user; 
         $p++; 
         } 
         return $marques[0]['tax_id']; 
         } 
         else return null; 
} 

所示的误差是:在此行$result = mysql_query($qry,$this->connection);Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/admin/public_html/hitlist/include/fg_membersite.php on line 427

是什么问题?我该如何解决它?

+1

你能尝试一下这样的mysql_query($ QRY);它表示$ this->连接没有引用有效的mysql连接链接资源 –

+1

检出与数据库建立连接的函数。另外'var_dump($ this-> connection)'以检查它的值。 –

+0

如何从phpMyAdmin知道主机 –

回答

1

好吧,也许无关,但我认为这需要固定

$qry = "insert into user_to_birds(user_id,tax_id)values(1 ,'.$id .') " 

$qry = "insert into user_to_birds(user_id,tax_id)values(1 ,".$id .") " 

$qry = "insert into user_to_birds(user_id,tax_id)values(1 ,$id) "