JSON + jQuery 不工作
我试图让 jQuery 获取 JSON 文件，并在按下按钮时将其中的数据放在一个简单的站点上。 因此，JSON 代码如下所示：
{
"images" : [
{ "source" = "images1", "alternative" = "altImg1" },
{ "source" = "images2", "alternative" = "altImg2" },
{ "source" = "images3", "alternative" = "altImg3" }
]
}
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>
Firebug 未检测到错误。我多次重写代码，查找错误，将其与类似代码进行比较等等，但找不到任何东西。
提前致谢！
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!
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评论（1）
您的 json 表示法是错误的，
请使用
:而不是=，例如：your json notation is wrong
use
:instead of=like: