jkwabla
06-23-2011, 08:28 AM
public function open_connection(){
$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS);
if(!$this->connection){
die("Database connection failed: ". mysql_error());
}
else{
$db_select = mysql_select_db(DB_NAME,$this->connection);
if(!$db_select){
die("Database connection failed: ". mysql_error());
}
}
}
do you find any error in my code ? i keep getting this error Warning: mysql_connect() [function.mysql-connect]: Access denied for user ...
$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS);
if(!$this->connection){
die("Database connection failed: ". mysql_error());
}
else{
$db_select = mysql_select_db(DB_NAME,$this->connection);
if(!$db_select){
die("Database connection failed: ". mysql_error());
}
}
}
do you find any error in my code ? i keep getting this error Warning: mysql_connect() [function.mysql-connect]: Access denied for user ...