CODEWITHSUNDEEP
phpcodeigniter
Anthony
Anthony

Reputation: 4050

Undefined JSON variable in CodeIgniter View

Why am I getting the following errors

A PHP Error was encountered Severity: Warning Message: Undefined variable: json Filename: views/search_page.php Line Number: 8

A PHP Error was encountered Severity: Warning Message: Trying to get property of non-object Filename: views/search_page.php Line Number: 8

A PHP Error was encountered Severity: Warning Message: Invalid argument supplied for foreach() Filename: views/search_page.php Line Number: 8

with this code?

search.php

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');

class Search extends CI_Controller {

    public function index()
    {

        $json = json_decode(file_get_contents('http://search.twitter.com/search.json?q=to%3astackexchange'));

        $this->load->view('search_page', $json);
    }
}

/* End of file search.php */
/* Location: ./application/controllers/search.php */

search_page.php

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <title>Twitter Test</title> 
</head>
<body>
<?php foreach ($json->results as $result): ?>
    <h2><?php echo $result->from_user; ?></h2>
<?php endforeach ?>
</body>
</html>

Upvotes: 1

Views: 967

Answers (1)

Madbreaks
Madbreaks

Reputation: 19539

You need to assign the variable you're passing in ($json) to a name ("json")

$this->load->view('search_page', array('json' => $json));

Perhaps a more clear example:

$this->load->view('search_page', array('myNeatObject' => $json));

// ...then, in your view, you could

<p>This is the JSON: <?php echo print_r($myNeatObject, true) ?></p>

That's how you name a variable for access in a view.

Upvotes: 4

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