This is an old post, but I'm posting a response anyway:
Let's assume you want to get the jSON code generated by the following file, "get_json_code.php":
<?php
$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo json_encode($arr);
?>
Like you mentioned, $.getJSON() uses JSONP when you add a "jsoncallback=?" parameter to the required URL's string. For example:
$.getJSON("http://mysite.com/get_json_code.php?jsoncallback=?", function(data){
alert(data);
});
However, in this case, you will get an "invalid label" message in Firebug because the "get_json_code.php" file doesn't provide a valid reference variable to hold the returned jSON string. To solve this, you need add the following code to the "get_json_code.php" file:
<?php
$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo $_GET['jsoncallback'].'('.json_encode($arr).')'; //assign resulting code to $_GET['jsoncallback].
?>
This way, the resulting JSON code will be added to the 'jsoncallback' GET variable.
In conclusion, the "jsoncallback=?" parameter in the $.getJSON() URL does two things: 1) it sets the function to use JSONP instead of JSON and 2) specifies the variable that will hold the JSON code retrieved from the "get_json_code.php" file. You only need to make sure they have the SAME NAME.
Hope that helps,
Vq.
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