Simple JSON request in PHP - php

Question

I have the following json
country_code({"latitude":"45.9390","longitude":"24.9811","zoom":6,"address":{"city":"-","country":"Romania","country_code":"RO","region":"-"}})
and i want just the country_code, how do i parse it?
I have this code
<?php
$json = "http://api.wipmania.com/jsonp?callback=jsonpCallback";
$jsonfile = file_get_contents($json);
var_dump(json_decode($jsonfile));
?>
and it returns NULL, why?
Thanks.

Answer 1

<?php
$jsonurl = "http://api.wipmania.com/json";
$json = file_get_contents($jsonurl);
var_dump(json_decode($json));
?>
You just need json not jsonp.
You can also try using json_decode($json, true) if you want to return the array.

Answer 2

you're requesting jsonp with http://api.wipmania.com/jsonp?callback=jsonpCallback, which returns a function containing JSON like:
jsonpCallback({"latitude":"44.9718","longitude":"-113.3405","zoom":3,"address":{"city":"-","country":"United States","country_code":"US","region":"-"}})
and not JSON itself. change your URL to http://api.wipmania.com/json to return pure JSON like:
{"latitude":"44.9718","longitude":"-113.3405","zoom":3,"address":{"city":"-","country":"United States","country_code":"US","region":"-"}}
notice the second chunk of code doesn't wrap the json in the jsonpCallback() function.

Answer 3

The website doesn't return pure JSON, but wrapped JSON. This is meant to be included as a script and will call a callback function. If you want to use it, you first need to remove the function call (the part until the first paranthesis and the paranthesis at the end).

Answer 4

If your server implements JSONP, it will assume the callback parameter to be a JSONP signal and the result will be similar to a JavaScript function, like
jsonpCallback("{yada: 'yada yada'}")
And then, json_decode won't be able to parse jsonpCallback("{yada: 'yada yada'}") as a valid JSON string

Answer 5

If country_code( along with closing parenthesis are include in your json, remove them.
This is not a valid json syntax: json

Answer 6

You are being returned JSONP, not JSON. JSONP is for cross-domain-requests in JavaScript. You don't need to use it when using PHP because you aren't affected by cross-domain-policies.
Since you are getting a string from the file_get_contents() function you can do a replacement of the country_code( text (this is the JSONP specific part of the response):
<?php
$json = "http://api.wipmania.com/jsonp?callback=jsonpCallback";
$jsonfile = substr(file_get_contents($json)), 13, -1);
var_dump(json_decode($jsonfile));
?>
Note
This works but JKirchartz's solution looks better, just request the correct data rather than messing around with the incorrect data.

Answer 7

Obviously in this situation, using the correct URL to access the API will return pure jSON.
"http://api.wipmania.com/json"
A lot of people are providing an alternative to the API in use, rather than answering the OP's question, so here is a solution for those looking for a way of handling jSONp in PHP.
First, the API allows you to specify a callback method, so you can either use Jasper's method of getting the jSON sub string, or you can give a callback method of json_decode, and modify the result to use with a call to eval. This is my alternative to Jasper's code example since I don't like to be a copy cat:
$json = "http://api.wipmania.com/jsonp?callback=json_decode";
$jsonfile eval(str_replace("(", "('", str_replace(")", "')", file_get_contents($json)))));
var_dump($jsonfile);
Admittedly this seems a little longer, more insecure, and not as clear to read as Jasper's code:
$json = "http://api.wipmania.com/jsonp?callback=jsonpCallback";
$jsonfile = substr(file_get_contents($json)), 13, -1);
var_dump(json_decode($jsonfile));
Then the jSON "address":{"city":"-","country":"Romania","country_code":"RO","region":"-"} tells us to access the country_code like so:
$jsonfile->{'address'}->{'country_code'};