[ borcha @ 16.07.2013. 17:53 ] @
Pokusavam izvuci parametre iz .INI fajla za povezivanje za bazom.. naziv fajla serset.ini i nalazi se u korenom dir..... sa nazivom INI (../INI/serset.ini). Code: [mysql] mysq_hostname=localhost mysq_user=root mysq_pass=root mysq_port=3306 mysq_dbname=baza napravio sam i class-u CON nalazi se u faljlu : condata.php . Code: class CON { public $host; public $user; public $pass; public $port; public $database; function __construct() { $pain= parse_ini_file("INI/serset.ini",true); $this->host=$pain['mysql']['mysq_hostname']; $this->user=$pain['mysql']['mysq_user']; $this->pass=$pain['mysql']['mysq_pass']; $this->port=$pain['mysql']['mysq_port']; $this->database=$pain['mysql']['mysq_dbname']; $this->openCon(); } public function openCon(){ $con=mysql_connect($this->host, $this->user, $this->pass); if(!$con){ die('Greska u povezivanju sa bazom: ' . mysql_error()); } mysql_select_db($this->database,$con); mysql_query("set names 'utf8'"); } } Desava se da ne moze pronaci fajl tj uraditi parsiranje: Code: Warning: parse_ini_file(INI/serset.ini) [function.parse-ini-file]: failed to open stream: No such file or directory in C:\xampp\htdocs\<SAJTFOLDER>\php\condata.php on line 15 |