Print out JSON with PHP - php

Question

Nothing shows on the screen, is this valid code below? I know theres a JSON parameter called 'text' within the received data but not sure how to print it out?
<?php
$url='http://twitter.com/statuses/user_timeline/twostepmedia.json'; //rss link for the twitter timeline
//print_r(get_data($url)); //dumps the content, you can manipulate as you wish to
$obj = json_decode($data);
print $obj->{'text'};
/* gets the data from a URL */
function get_data($url)
{
$ch = curl_init();
$timeout = 5;
curl_setopt($ch,CURLOPT_URL,$url);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout);
$data = curl_exec($ch);
curl_close($ch);
return $data;
}
?>

Answer 1

This should work:
$obj = json_decode(get_data($url));
$text = $obj[0]->text;
It's a good idea to try something like var_dump($obj) when you run into an issue like this. After doing so, it becomes immediately clear $obj[0]->text is what you're after.
#benhowdle89 comment:
foreach ($obj as $item) {
$text = $item->text;
}

Answer 2

You should assign the value returned by get_data to a variable and pass it to json_decode i.e.:
<?php
$url='http://twitter.com/statuses/user_timeline/twostepmedia.json'; //rss link for the twitter timeline
//print_r(get_data($url)); //dumps the content, you can manipulate as you wish to
$data = get_data($url);
$obj = json_decode($data);
print $obj->text;
/* gets the data from a URL */
function get_data($url)
{
$ch = curl_init();
$timeout = 5;
curl_setopt($ch,CURLOPT_URL,$url);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout);
$data = curl_exec($ch);
curl_close($ch);
return $data;
}
?>

Answer 3

$data is not set, and you don't need the curly braces.