如何将更多参数传递给该函数？

I have this function:

function mobio_checkcode($servID, $code, $debug=0) {

$res_lines = file("http://www.mobio.bg/code/checkcode.php?servID=$servID&code=$code");

$ret = 0;
if($res_lines) {

    if(strstr("PAYBG=OK", $res_lines[0])) {
        $ret = 1;
    }else{
        if($debug)
            echo $line."
";
    }
}else{
    if($debug)
        echo "Unable to connect to mobio.bg server.
";
    $ret = 0;
}

return $ret;
}

And here how i use it:

if(mobio_checkcode($servID, $code, 0) == 1) {
 echo "Code is valid!!";
}


$code = $_REQUEST['code'];
$servID = 29;
$post = $_REQUEST['post'];

Here i have form! In this form u enter code and display is valid or no so i want to pass more $code like

$code1 = $_REQUEST['code1'];
$code2 = $_REQUEST['code2'];
$code3 = $_REQUEST['code3'];

I want to pass more variables to function how it will be done.. Please help me thank u <3

Just pass all of variables as array like

$array = ['servID'=>29,'code'=>XXXX];

function checkcode($array) {
//Do stuff
$array['servID']
$array['code']
}

function mobio_checkcode($servID, $code, $debug=0, $element1, $element2) {

$res_lines = file("http://www.mobio.bg/code/checkcode.php?servID=$servID&code=$code");

$ret = 0; if($res_lines) {

if(strstr("PAYBG=OK", $res_lines[0])) {
    $ret = 1;
}else{
    if($debug)
        echo $line."
";
}

}else{ if($debug) echo "Unable to connect to mobio.bg server. "; $ret = 0; }

return $ret;

//you can pass extra elements so on.. otherwise create an array of elements and pass only array to function