Twitter Trends PHP - php

Question

I am trying to use http://api.twitter.com/1/trends/current.json?exclude=hashtags however I am having some trouble.
So, I'm trying to use:
<?php
$init = 'http://api.twitter.com/1/trends/current.json?exclude=hashtags';
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,$init);
curl_setopt($ch, CURLOPT_RETURNTRANSFER,1);
$result = curl_exec($ch);
curl_close($ch);
$obj = json_decode($result, true);
foreach ($obj[0]['trends'] as $trend) {
print $trend['query'];
echo "<br>";
print $trend['name'];
echo "<hr>";
}
?>
And I am getting this Error:
Notice: Undefined offset: 0 in \index.php on line 11
Warning: Invalid argument supplied for foreach() \index.php on line 11

Answer 1

You read the JSON wrong. You need to do something like the following:
foreach ($obj['trends']['2011-02-23 18:00:00'] as $trend) {
print $trend['query'];
echo "<br>";
print $trend['name'];
echo "<hr>";
}

Answer 2

If I am not wrong
$obj = json_decode($result, true);
would produce an array and not an object. secondly using the twitter api is as simple as:
<?php
function get_trends($woeid){
return json_decode(file_get_contents("http://api.twitter.com/1/trends/".$woeid.".json?exclude=hashtags", true), false);
}
$data = get_trends(23424848); //23424848 is woeid for India...
$trends = $data[0]->trends;
echo "<ul>";
if(!empty($trends)){
foreach($trends as $trend){
echo '<li>'.$trend->name.'</li>';
}
}
echo "</ul>";
?>