JSON + jQuery not working

Question

I'm trying to make jQuery take JSON file and put the data from it on a simple site, when a button is pressed. So, the JSON code looks like this:

{
    "images" : [
        { "source" = "images1", "alternative" = "altImg1" },
        { "source" = "images2", "alternative" = "altImg2" },
        { "source" = "images3", "alternative" = "altImg3" }
    ]
}

And the HTML + jQuery:

<html xmlns="http://www.w3.org/1999/xhtml">
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
        "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <head>
        <title>jQuery</title>
        <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>  
    </head>

    <body>
        <button>Press Me!</button>
        <script>
            $('button').click(function()    {
                $.getJSON('json-db.html', function(data)    {
                    for(var i = 0; i < data.images.length; i++) {
                        var image = data.images[i];
                        $('#result').append('<h1>' + image.source + ' ' + image.alternative + '</h1>');
                    }
                });
            });
        </script>
        <div id="result">Result</div>
    </body>
</html>

There are no errors detected by Firebug. I rewrote the code several times, looked for mistakes, compared it to a similar code and so on, but couldn't find anything.

Thanks in advance!

Caspar Kleijne · Accepted Answer · 2011-04-23 08:02:41Z

3

your json notation is wrong

use : instead of = like:

..........
"images" : [
    { "source" : "images1", "alternative" : "altImg1" },
    ....................
]
..........

answered Apr 23, 2011 at 8:02

Caspar Kleijne

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3 Comments

Tsvetan Over a year ago

Oh, how dumb am I... Thanks a lot!

Caspar Kleijne Over a year ago

I guess jQuery gracefully cloackes the error. Try using a json validator to pick out small future errors jsonlint.com ;)

Tsvetan Over a year ago

Yeah, I don't like web developing because of this(and the amount of code you need to write). Thanks again!

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