Variables not passing to class (PHP)

Question

I am trying to pass a JSON response from an external API to a class in a different file. Somehow it's not storing in the class. I've tried var_dumping the class like this:

$user->country_ip = $country_ip; die(var_dump($user->country_ip));

It somehow doesn't work. But, when I try to vardump the actual variable like this:

die(var_dump($country_ip));

Then it does work, so the variable is fine. I'm very confused, does anyone have an answer to this? It's midnight here, it might be something small?

What do you mean? The variable itself is global and accessible throughout the entire file — Sebastian Boender
– Sebastian Boender, Commented Jul 16, 2019 at 23:45
Can you be more specific about what "doesn't work" means? I used that same code and it dumped the value as expected. — Don't Panic
– Don't Panic, Commented Jul 17, 2019 at 0:01

Vinnie · Accepted Answer · 2019-07-17 00:20:12Z

1

A minimalist example will be like this,

class User
{
  public $country_ip;

  public function getCountryIp()
  {
    return json_decode($this->country_ip, true);
  }
}

$country_ip = '192.168.2.227';
$user = new User();
$user->country_ip= json_encode($country_ip);
var_dump($user->getCountryIp()); //It will print "192.168.2.227"

DEMO: https://3v4l.org/YSk9X

edited Jul 17, 2019 at 0:20

Vinnie

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answered Jul 16, 2019 at 23:46

A l w a y s S u n n y

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2 Comments

Sebastian Boender Over a year ago

The function is not within the class, it's actually in a different file. Does that make a difference? I have obviously created the class, etc. And it works with all my other variables which I post towards the file. It just doesn't seem to work with this variable from a new function?

Sebastian Boender Over a year ago

Nvm! Your answer worked! Thanks a lot man, I just overlooked the public part and forgot to do it for the country_ip variable

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