YQL JSON脚本没有返回？ [英] YQL JSON script not returning?
本文介绍了YQL JSON脚本没有返回？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我更直接关闭<一这里有一个脚本，复制pretty的href="http://net.tutsplus.com/tutorials/javascript-ajax/quick-tip-cross-domain-ajax-request-with-yql-and-jquery/"相对=nofollow>这个。为什么不code，下面列出，返回什么？
I have a script here, copied pretty much directly off this. Why doesn't the code, listed below, return anything?
ajax.html ：
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
"http://www.w3.org/TR/html4/strict.dtd">
<html dir="ltr" lang="en-US">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Cross-Domain Ajax Demo</title>
</head>
<body>
<div id="container">
<form>
<p><label>Type a URL:</label><input type="text" name="sitename" id="sitename"/></p>
<p><input type="submit" name="submit" id="submit" value="Make Cross Domain Ajax request"</p>
</form>
</div>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js" charset="utf-8"></script>
<script type="text/javascript" src="cross-domain-requests.js"></script>
<script type="text/javascript">
$('form').submit(function() {
var path = "www.google.com";
requestCrossDomain(path, function(results) {
$('#container').html(results);
});
return false;
});
</script>
</body>
</html>
跨域-requests.js ：
// Accepts a URL and a callback function to run.
function requestCrossDomain( site, callback ) {
// If no URL was passed, exit.
if ( !site ) {
alert('No site was passed.');
return false;
}
// Take the provided URL, and add it to a YQL query. Make sure you encode it!
var yql = 'http://query.yahooapis.com/v1/public/yql?q=' + encodeURIComponent('select * from html where url="' + site + '"') + '&format=xml&callback=cbFunc';
// Request that YSQL string, and run a callback function.
// Pass a defined function to prevent cache-busting.
$.getJSON( yql, cbFunc );
function cbFunc(data) {
// If we have something to work with...
if ( data.results[0] ) {
// Strip out all script tags, for security reasons.
// BE VERY CAREFUL. This helps, but we should do more.
data = data.results[0].replace(/<script[^>]*>[\s\S]*?<\/script>/gi, '');
// If the user passed a callback, and it
// is a function, call it, and send through the data var.
if ( typeof callback === 'function') {
callback(data);
}
}
// Else, maybe we requested a site that doesn't exist, and nothing returned.
else throw new Error('Nothing returned from getJSON.');
}
}
（我是比较新的脚本和Ajax，所以我提前道歉，如果我做任何愚蠢的事。）
(I'm relatively new to scripting and Ajax, so I apologise in advance if I do anything stupid.)
推荐答案
试着改变回调VAR YQL回调=？和SELECT语句从XML'是这样的：
Try changing the callback in var yql to callback=? and the select statement to 'from xml' like this:
var yql = 'http://query.yahooapis.com/v1/public/yql?q=' + encodeURIComponent('select * from xml where url="' + site + '"') + '&format=xml&callback=?';
这篇关于YQL JSON脚本没有返回？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
查看全文
