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I'm pretty sure it's an obvious error somewhere on this - but let me explain what I'm donig :

I'm calling a PHP file via jQuery as follows :

$.getJSON("/phpincs/getbucket.php?awskey="+awskey+"&awssecret="+awssecret+"&bucket="+bucket,
    function(json){
        $.each(json,function(i,item){
        $(new Option(item.name,item.name)).appendTo('#bucket_contents_0');
        });
    }

and the JSON file it returns is as follows :

Array
(
    [bt_shop.png] => Array
        (
            [name] => bt_shop.png
            [time] => 1260393948
            [size] => 156985
            [hash] => 8a4eba621d5e08b84c962a0ad21ec2ae
        )

    [me_profile.jpg] => Array
        (
            [name] => me_profile.jpg
            [time] => 1260393952
            [size] => 2714
            [hash] => 4f5d185b0c671e6165902762681f098b
        )

    [thumbnail.aspx.jpeg] => Array
        (
            [name] => thumbnail.aspx.jpeg
            [time] => 1260393951
            [size] => 5268
            [hash] => 6939efa58ff7ffcac17218ffdd0d5d8c
        )

)
true

for some reason it doesn't seem to fire the function(json){} - I've stuck an alert(''); in and it doesn't do anything.

Can someone quickly explain to me what seems to be going wrong there?

Cheers,

Carl

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4
  • 3
    The "JSON file" you say is returned is absolutely not JSON ; is this really what your PHP script is returning ? If yes, try returning "real" JSON instead (see the PHP json_encode function) Commented Dec 9, 2009 at 23:14
  • 1
    Agreed. That PHP output is not JSON. Use jsonlint.com to verify you have valid JSON before trying to do something with it. Commented Dec 9, 2009 at 23:18
  • you beauty! of Course it's not!!! I'm sorry, It's 11:30pm and I've been working on one thing or another since 8:30 this morning.... :) Many Thanks Pascal. Commented Dec 9, 2009 at 23:18
  • @Carl : you're welcome :-) I thought "it can't be it" s didn't make an answer of it ; should have had ^^ ; Have fun for the rest of the... day/night ^^ Commented Dec 9, 2009 at 23:22

4 Answers 4

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4

It is more than likely not calling the callback function because it doesn't look like what you are returning is json. If you have an $variable defined that contains your array...call 

echo json_encode($jsondata); exit;

At the end of your script.

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3

I've changed the names of the inner arrays as your previous labels are going to cause problems with the dot. You will get an error like:

myArray.bt_shop is undefined

when you try to call

alert(myArray.bt_shop.png.name);

the only way it could be called is with

alert(myArray["bt_shop.png"].name);

So having changed your code a bit, this is the JSON version of your arrays...

{
    "one":
    {
        "name": "bt_shop.png",
        "time": "1260393948",
        "size": "156985",
        "hash": "8a4eba621d5e08b84c962a0ad21ec2ae"
    },

    "two":
    {
        "name": "me_profile.jpg",
        "time": "1260393952",
        "size": "2714",
        "hash": "4f5d185b0c671e6165902762681f098b"
    },

    "three":
    {
        "name": "thumbnail.aspx.jpeg",
        "time": "1260393951",
        "size": "5268",
        "hash": "6939efa58ff7ffcac17218ffdd0d5d8c"
    }
}

Then you reference your fields like this when you have the object:

myArray["two"]["name"]
myArray["two"].name
myArray.two.name
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Comments

1

Your returned file isn't JSON. Unless you're using PHP syntax to describe your JSON object for us, you need to encode it into JSON format using json_encode.

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1

What you call the JSON file isn't JSON. Or maybe do you use some PHP library that converts that strange format into JSON ?

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