[jQuery] Re: Problems with getJSON between browsers

[Kus]

Hi,

I to was playing with the Google Latitude Badge API.

For some reason jQuery doesnt like it, so what i did was compare what
Google returns to what the Flickr example does

http://api.flickr.com/services/feeds/photos_public.gne?tags=cat&tagmode=any&format=json&jsoncallback=?

The flickr one is wrapped in brackets ( ), it appears you need these
when using the $.getJSON function.

This does not fix it yet, the format was pretty dodgy, so what i did
was call my own script (with the user ids i wanted ?user=<ids>) and
the curl the Google Latitude Badge API into a variable.

Then parse the variable through json_decode into an array, then parse
that through json_encode so its back into a single variable and has
been cleaned.

I then did a check to see if the data is encapsulated in brackets ( ),
if not (which it isnt at them moment) i add them.

Then output the application/x-javascript content-type header and echo
the cleaned json variable.

I can now call my script via window.location.href+"?user=1234" which
will output cleaned json encapsulated in brackets.

PHP CODE:

<code>
<?php
if(!empty($_GET['user'])){
        $ch = curl_init();
        curl_setopt($ch, CURLOPT_URL, "http://www.google.com/latitude/apps/
badge/api?user=".$_GET['user']."&type=json");
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
        curl_setopt($ch, CURLOPT_TIMEOUT, 120);
        $response = curl_exec($ch);
        if(curl_errno($ch)){
            echo curl_error($ch);
        }else{
                $json = json_encode(json_decode($response));
                if(substr($json,0,1) != "(") $json = "(".$json;
                if(substr($json,-1) != ")") $json = $json.")";
                header('Content-Type: application/x-javascript; charset=utf-8');
                echo $json;
        }
        curl_close($ch);
}else{
?>
--- your html ---
<?php
}
?>
</code>

So you think this would work in $.getJSON now, it doesnt, i dont know
why. So instead i load it via $.get and eval() the data which works.

JAVASCRIPT:

<code>
$.get(window.location.href+"?user=1234", function(data){
        $.each(eval(data), function(i,item){
                alert(i+"=>"+item);
        });
});
</code>

Hope this helps people!

On Jul 29, 12:29 am, yaphi <jconto...@gmail.com> wrote:
> Hello,
>
> I am trying to request the json feed from Google Latitude. This script
> is working perfectly in Safari 4, but I get a 400 Bad Request error in
> FF 3.5. If I add the &callback=? to the end of the URL, then FF 3.5
> hits me with an Invalid Label error which I can't work around (tried
> eval). Also, adding &callback=? to the end of the URL results in a
> syntax error in result in Safari 4.
>
> Can anyone point me in the right direction here?
>
> <code>
> <script src="http://www.google.com/jsapi";></script>
> <script type="text/javascript" charset="utf-8">
>
>         // Just using the google ajax library service for simplicity sake.
> Replace this with the regular jQuery installation
>
>         google.load("jquery", "1");
>         google.setOnLoadCallback(function() {
>                 loadLatitudeLocationData();
>         });
>
>         function loadLatitudeLocationData()
>         {
>                 var url = "http://www.google.com/latitude/apps/badge/api?
> user=1798529584061495106&type=json";
>                 $.getJSON(url,
>                 function(data) {
>                         var features = data.features;
>                         console.log(features[0].properties.reverseGeocode);
>                 });
>         }
> </script>
> </code>