Question

So I’m playing around with yahoo’s yql. I got it to generate a url and the url looks like it’s returning decoded json to me. If I attempt to decode the output of the yql url i get no results (so assuming i’m right there)

$c =curl_init("http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20local.search%20where%20state%3D'delaware'%20and%20city%20%3D%20'smyrna'%20and%20query%3D'pizza'&format=json"); 

curl_setopt($c, CURLOPT_RETURNTRANSFER, true);
curl_setopt($c, CURLOPT_CONNECTTIMEOUT, 20); // query times out after 20 seconds

$data = curl_exec($c); // I asked for data format to be in json in the query it appears to be returned decoded

curl_close($c);
    //print_r($data);
    //$arr[] = $data; (returning results in decoded json)
    //print_r($arr); (returning results)
foreach($data->query->results->result as $result)
{
    echo 'blah blah blah';
}

However, no matter how I do things I can’t seem to access the output in my foreach line. What am I doing wrong?

Thanks in advance

Answer 1

You have to decode the json string first, otherwise it remains a plain-text string:

$data = curl_exec($c);
if ($data === FALSE) {
    die("Curl failed with error: " . curl_error($c));
}
$data = json_decode($data);
if (is_null($data)) {
    die("json_decode failed with error: " . json_last_error());
}

foreach(...) {
}

Note I’ve added error handling – your code had none, and assumed that the curl call succeeded. This is a bad way to go, as you can NOT depend on external resources to present or functional. Always check if an external resource request succeeded before proceeding.