JSON output for mapping data using json_decode() in php - php

Question

I am attempting to write PHP code to interact with JSON output from Mapquest's Open API / Open Street Map service. I have listed it below. I have been using this code in my Drupal 6 implementation. This code returns no output. When I use it, json_last_error() outputs 0.
function json_test_page() {
$url = 'http://open.mapquestapi.com/directions/v1/route?outFormat=json&from=40.037661,-76.305977&to=39.962532,-76.728099';
$json = file_get_contents($url);
$obj = json_decode(var_export($json));
$foo .= $obj->{'fuelUsed'};
$output .= foo;
return $output;
}
You can view the raw JSON output by following the URL. In this function I am expecting to get 1.257899 as my output. I have two questions:
(1) What can I call so I get items out of my array. For instance, how can I get the value represented in JSON "distance":26.923 out of the array?
(2) Is it possible am I running into a recursion limit issue that I've read about in the PHP Manual?

Answer 1

If you read the manual page for json_decode carefully, you'll notice there is a parameter (false by default) that you can pass to have it return an array rather than an object.
$obj = json_decode($json, true);
So:
<?php
function json_test_page() {
$url = 'http://open.mapquestapi.com/directions/v1/route?outFormat=json&from=40.037661,-76.305977&to=39.962532,-76.728099';
$json = file_get_contents($url);
$obj = json_decode($json, true);
//var_dump($obj);
echo $obj['route']['fuelUsed'];
}
json_test_page();

Answer 2

Remove the var_export function from json_decode.
You're trying to convert information about a string to json.
I was able to get the fuelUsed property this way
function json_test_page() {
$url = 'http://open.mapquestapi.com/directions/v1/route?outFormat=json&from=40.037661,-76.305977&to=39.962532,-76.728099';
$json = file_get_contents($url);
$obj = json_decode($json);
return $obj->route->fuelUsed;
}