get results returned in a function - php

Question

i have this inside a php function:
$result = new stdClass();
$result->domainname = $domainName;
$result->element = $element;
$result->availability = $availability;
return ($result);
so its returning all of the values in the $result variable
when i do a print_r on the function, the results display like this:
stdClass Object
(
[domainname] => domain.com
[element] =>
[availability] => false
)
i am calling the function with this code:
$domain = domainNameCheck($_GET["domain"].'.'.$_GET["tld"]);
so i tried to get the returned by doing $domain->availability but its not returning the value, example:
if($domain->availability) {
echo 'yes';
} else {
echo 'no';
}
am i trying to get the data the incorrect way?
UPDATE
the full function is:
if(!function_exists("domainNameCheck")) {
function domainNameCheck($domainName, $element) {
$result = '';
$client = new IcukApiClient();
$client->username = "username";
$client->key = "pass";
$client->encryption = "SHA-512";
$req = new IcukApiRequest();
$req->url = "/domain/availability/" . $domainName;
$req->method = "GET";
$res = $client->send($req);
$availability = 'unknown';
if ($res->success) {
$obj = json_decode($res->response);
$availability =($obj->available) ? 'true' : 'false';
}
else {
$availability = 'unknown';
}
$result = new stdClass();
$result->domainname = $domainName;
$result->element = $element;
$result->availability = $availability;
return ($result);
}
}

Answer 1

Your main problem seems to be that you are calling a function with 2 parameters but passing only one parameter
function domainNameCheck($domainName, $element) {}
// called like this (one parameter)
$domain = domainNameCheck($_GET["domain"].'.'.$_GET["tld"]);
This should be generating a compile error!
Also here
if ($res->success) {
$obj = json_decode($res->response);
// check what $obj->available is set to
// it may also be a string and not a boolean
print_r($obj);
$availability =($obj->available) ? 'true' : 'false';
}
else {
$availability = 'unknown';
}

Answer 2

Please note that there are two error/warning messages PHP is giving:
E_WARNING : type 2 -- Missing argument 2 for domainNameCheck()
E_NOTICE : type 8 -- Undefined variable: element
You should fix those errors, and make sure you are informed of errors during development.
Secondly, you have defined your availability as a string by assigning "false", "true", or "unknown". So when you do this:
if($domain->availability) {
... that will be true for all three values, because strings are true for PHP when converted to boolean (except when empty). To illustrate, this will echo "hello":
if ("false") echo "hello";
So you need to change your test like this:
if($domain->availability === "true") {
Or, If you want to define $domain->availability as a true boolean, then you need to alter the assignments in your function, like this:
....
$availability = $obj->available; // assuming that is a boolean!
}
else {
$availability = null; // unknown
}
... and then you can do what you had:
if($domain->availability) {

Answer 3

Likely because $domain->availability is boolean
To output you can first check whether its true or false and output accordingly
here's a simple example:
if ($domain->availability){
echo 'Available';
}
else {
echo 'Not Available';
}