PHP: Handling 'JSONP' output vs 'JSON', and its parsing?

Question

I am having a problem parsing 'jsonp' request with php's json_decode function.

My questions is

a. What is the use of call back function in 'jsonp', should i just trip that off, or am I suppose to use it in some manner. ?

b. How can I rectify the syntax error received in 'jsonp' format ?

Below I have given the code and the response that I get.

1. I request a sample url with PHP's curl

$url = 'https://ssl.domain.com/data/4564/d.jsonp';

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);                
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);        
curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 5);        
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE);    
curl_setopt($ch, CURLOPT_USERAGENT, "Mozilla/4.0 (compatible; MSIE 5.01; Windows NT 5.0)");     
$feed = curl_exec($ch);
curl_close($ch);

echo $feed =  gzdecode($feed); // Success its displays the jsonp feed

2. Then I try to json_decode the received output, which throws the error no 4 meaning JSON_SYNTAX_ERROR, the reason I guess is because names of string type in jsonp are not quoted. e.g. Categories, Name, Position etc.

$json_feed = json_decode($feed);
$error = json_last_error(); 
echo $error;     // Throws error no. 4

3. RAW 'jsonp' output from the url.

domain_jsonp_callback({
   Categories:[
      {
         Name:"Artifacts",
         Position:14,
         Count:70,
         ImageUrls:{
            i100:"//s3-eu-west-1.amazonaws.com/s.domain.com/1.png",
            i120:"//s3-eu-west-1.amazonaws.com/s.domain.com/2.png",
            i140:"//s3-eu-west-1.amazonaws.com/s.domain.com/3.png",
            i180:"//s3-eu-west-1.amazonaws.com/s.domain.com/4.png",
            i220:"//s3-eu-west-1.amazonaws.com/s.domain.com/5.png",
            i280:"//s3-eu-west-1.amazonaws.com/s.domain.com/6.png"
         }
      }
   ]
});

Answer 1

What is the use of call back function in 'jsonp', should i just trip that off, or am I suppose to use it in some manner. ?

JSON-P is really a JavaScript script that consists of a function call with an argument.

If you want to parse it in PHP, then yes, you need to strip it off. You also need to strip off the ); at the end.

b. How can I rectify the syntax error received in 'jsonp' format ?

You need to fix the data so it really is JSON. The data you have is a JavaScript literal, but it doesn't conform to the subset of JavaScript that matches JSON (e.g. property names are not strings but must be).

It would be better to get a real JSON resource form the source instead.