Error in creating JSON in PHP - php

Question

I'm very new to PHP and JSON, so I hope you'll excuse my (probably) stupid question.
I'm trying to create a new JSON with fundamental information for my needings, using the information returned by Google Maps API.
Here's the code:
<?php
/**
* Created by PhpStorm.
* User: biagiomontesano
* Date: 22/10/15
* Time: 18:16
*/
function selectLongestSteps($steps_txt, $min_meters)
{
$steps = JSON.parse($steps_txt);
//$steps = json_decode($steps_txt, true);
//$steps = $steps_txt;
echo 'Num of steps: ' . count($steps[0]);
}
function createJsonFromResponse($response_json)
{
// output json
$output_json = '[';
// number of steps
$num_steps = count($response_json['routes'][0]['legs'][0]['steps']);
//echo $num_steps;
// fill the json
for($i = 0; $i<$num_steps; $i++)
{
// start parenthesis
$output_json .= '{';
// start latitude
$output_json .= '"start_lat":' . $response_json['routes'][0]['legs'][0]['steps'][$i]['start_location']['lat'] . ',';
// start longitude
$output_json .= '"start_lng":' . $response_json['routes'][0]['legs'][0]['steps'][$i]['start_location']['lng'] . ',';
// end latitude
$output_json .= '"end_lat":' . $response_json['routes'][0]['legs'][0]['steps'][$i]['end_location']['lat'] . ',';
// end latitude
$output_json .= '"end_lng":' . $response_json['routes'][0]['legs'][0]['steps'][$i]['end_location']['lng'] . ',';
// step length
$output_json .= '"step_length":' . $response_json['routes'][0]['legs'][0]['steps'][$i]['distance']['value']. ',';
// html instruction
$output_json .= '"instruction":"' . $response_json['routes'][0]['legs'][0]['steps'][$i]['html_instructions'] . '"';
// closure parenthesis
$output_json .= '}';
// insert comma if required
if($i != $num_steps-1)
$output_json .= ',';
}
$output_json .= ']';
return $output_json;
}
function get_driving_information($start, $finish)
{
if (strcmp($start, $finish) != 0) {
$start = urlencode($start);
$finish = urlencode($finish);
$url = 'http://maps.googleapis.com/maps/api/directions/json?origin=' . $start . '&destination=' . $finish . '&sensor=false';
// get the json response
$resp_json = file_get_contents($url);
// decode the json
$resp = json_decode($resp_json, true);
return $resp;
}
else
return null;
}
try
{
$info = get_driving_information('via Tiburtina 538, Roma', 'via Ariosto 25, Roma');
$steps = null;
if(!$info)
echo 'No info';
else
$steps = createJsonFromResponse($info);
selectLongestSteps($steps, 200);
//echo $steps;
}
catch(Exception $e)
{
echo 'Caught exception: '.$e->getMessage()."\n";
}
As a test, I asked my function selectLongestSteps to display the number of elements in the JSON.
As you see, I tried to initialise variable $steps in 3 different ways: the last two return a wrong results (0 or 1, while result should be 18).
The first one, that I found on web, returns the following error:
Notice: Use of undefined constant JSON - assumed 'JSON' in /Applications/XAMPP/xamppfiles/htdocs/prova_php/create_json.php on line 11
Fatal error: Call to undefined function parse() in /Applications/XAMPP/xamppfiles/htdocs/prova_php/create_json.php on line 11
I suspect the problem is in creating JSON inside the other function.
Can you help?
Thanks

Answer 1

The problem is this line:
$steps = JSON.parse($steps_txt);
JSON.parse() ist a JavaScript function. You need something like this in PHP
$steps = json_decode($steps_txt);
http://php.net/manual/en/function.json-decode.php

Answer 2

PHP has in-built function to change array to json and vice-verse.
json_encode($array) will give json form of array.
json_decode($json, true) will give your array back from json. If you didn't give true as your second param, you will be getting an object instead of an array.