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我是php的初學者。和我有此SQL問題:php函數中的MySQL-Link錯誤
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 $marques[0]['tax_id'];
}
else return null;
}
所示的誤差是:在此行$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
。
是什麼問題?我該如何解決它?
你能嘗試一下這樣的mysql_query($ QRY);它表示$ this->連接沒有引用有效的mysql連接鏈接資源 –
檢出與數據庫建立連接的函數。另外'var_dump($ this-> connection)'以檢查它的值。 –
如何從phpMyAdmin知道主機 –