How to output variable inside an object in PHP

Question

I'm trying to use a variable from inside an object in PHP.

I've tried to access the variable like $object->json_output but I'm receiving undefined property errors. I'm expecting to regex this output and extract data which I'll use later on.

My code is:

class curl
  {
     public function curlPut($url, $JSON, $token)
     {
        $ch = curl_init($url);
        $popt = array(
           CURLOPT_CUSTOMREQUEST => 'PUT',
           CURLOPT_RETURNTRANSFER => TRUE,
           CURLOPT_SSL_VERIFYPEER => false,
           CURLOPT_POSTFIELDS => $JSON,
           CURLOPT_HTTPHEADER => array(
              'Content-Type: application/json',
              'Authorization:'.$token.''
           ));
        curl_setopt_array($ch, $popt);
        $json_output = curl_exec($ch);
        curl_close($ch);
        return var_dump($json_output);
     }
  };

$object1 = new curl;

$object1->curlPut($url, $JSON, $token);

preg_match_all('/"id":"([0-9]*)/', $object1->json_output, $idtest);
$id_array[] = array(
  'id' => $idtest[1]
);

where $json_output is the variable I need to access and $id_array is an array of IDs I need regexed from $json_output. How would I access $json_output to be used in my preg_match_all function?

I'm new to using class/objects so apologies if this is a silly question.

Any comments would be greatly appreciated!

Sam

Instead of `$json_output = ...` you would need to set `$this->json_output = ...`. — Paul, Apr 28 '16 at 13:39
Is there any reason you're using a regular expression to read a value from JSON, rather than using json_decode()? — iainn, Apr 28 '16 at 13:41
the common ans is http://stackoverflow.com/questions/4361553/php-public-private-protected — Mohammedshafeek C S, Apr 28 '16 at 13:44
@iainn. I'm looping through an array of JSON of variable length so I can't reliably extract what I need using json_decode. — Sam Comber, Apr 28 '16 at 13:54
@MohammedShafeek thank-you, I'll familiarize myself with that post — Sam Comber, Apr 28 '16 at 13:55

Sebastian Brosch · Accepted Answer · 2016-04-28T13:49:22.880

You have to set a property on your class like the following:

class Test {
  public $json_output = 'Test';
}

$test = new Test();
echo $test->json_output; // output: Test;

The property have to be public not private or protected to access outside the class.

Your Code should look like the following:

class curl {
  public $json_output = '';

  public function curlPut($url, $JSON, $token) {
    $ch = curl_init($url);
    $popt = array(
      CURLOPT_CUSTOMREQUEST => 'PUT',
      CURLOPT_RETURNTRANSFER => TRUE,
      CURLOPT_SSL_VERIFYPEER => false,
      CURLOPT_POSTFIELDS => $JSON,
      CURLOPT_HTTPHEADER => array(
        'Content-Type: application/json',
        'Authorization:'.$token.''
      ));
    curl_setopt_array($ch, $popt);
    $this->json_output = curl_exec($ch);
    curl_close($ch);
  }
}


$object1 = new curl();
$object1->curlPut($url, $JSON, $token);

preg_match_all('/"id":"([0-9]*)/', $object1->json_output, $idtest);       
$id_array[] = array('id' => $idtest[1]);

score 1 · Answer 2 · answered Apr 28 '16 at 13:38

Like this. You create class variable and then you set it using $this->json_output. After that, it is accessible via $object->json_output.

class curl
  {
     public $json_output;                            //Added by tilz0R
     public function curlPut($url, $JSON, $token)
     {
        $ch = curl_init($url);
        $popt = array(
           CURLOPT_CUSTOMREQUEST => 'PUT',
           CURLOPT_RETURNTRANSFER => TRUE,
           CURLOPT_SSL_VERIFYPEER => false,
           CURLOPT_POSTFIELDS => $JSON,
           CURLOPT_HTTPHEADER => array(
              'Content-Type: application/json',
              'Authorization:'.$token.''
           ));
        curl_setopt_array($ch, $popt);
        $this->json_output = curl_exec($ch);         //Edit by tilz0R $this-> added
        curl_close($ch);
        return var_dump($json_output);
     }
  };

How to output variable inside an object in PHP

2 Answers2