使用 javascript 获取链接标签的内容 - 而不是 CSS
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Get contents of link tag with javascript - not CSS
提问by sproketboy
Assuming I have
假设我有
<LINK rel="Index" href="index.html">
<LINK rel="Next" href="Chapter3.html">
<LINK rel="Prev" href="Chapter1.html">
(taken from w3 web site sample)
（取自 w3 网站示例）
Does anyone know if these are accessible through the JavaScript DOM?
有谁知道这些是否可以通过 JavaScript DOM 访问？
I want to know If I have link tags like this in the HTML document whether they are read like the main document and added to the DOM and if I can access their DOMs as well.
我想知道如果我在 HTML 文档中有这样的链接标签，它们是否像主文档一样被读取并添加到 DOM 中，以及我是否也可以访问它们的 DOM。
回答by otaxige_aol
I have this code:
我有这个代码：
<script type="text/javascript">
var your_url = 'http://www.example.com';
</script>
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js" ></script>
<script type="text/javascript">
// jquery.xdomainajax.js ------ from padolsey
jQuery.ajax = (function(_ajax){
var protocol = location.protocol,
hostname = location.hostname,
exRegex = RegExp(protocol + '//' + hostname),
YQL = 'http' + (/^https/.test(protocol)?'s':'') + '://query.yahooapis.com/v1/public/yql?callback=?',
query = 'select * from html where url="{URL}" and xpath="*"';
function isExternal(url) {
return !exRegex.test(url) && /:\/\//.test(url);
}
return function(o) {
var url = o.url;
if ( /get/i.test(o.type) && !/json/i.test(o.dataType) && isExternal(url) ) {
// Manipulate options so that JSONP-x request is made to YQL
o.url = YQL;
o.dataType = 'json';
o.data = {
q: query.replace(
'{URL}',
url + (o.data ?
(/\?/.test(url) ? '&' : '?') + jQuery.param(o.data)
: '')
),
format: 'xml'
};
// Since it's a JSONP request
// complete === success
if (!o.success && o.complete) {
o.success = o.complete;
delete o.complete;
}
o.success = (function(_success){
return function(data) {
if (_success) {
// Fake XHR callback.
_success.call(this, {
responseText: data.results[0]
// YQL screws with <script>s
// Get rid of them
.replace(/<script[^>]+?\/>|<script(.|\s)*?\/script>/gi, '')
}, 'success');
}
};
})(o.success);
}
return _ajax.apply(this, arguments);
};
})(jQuery.ajax);
$.ajax({
url: your_url,
type: 'GET',
success: function(res) {
var text = res.responseText;
// then you can manipulate your text as you wish
alert(text);
}
});
</script>
回答by Zoltan Toth
Try this - http://jsfiddle.net/TpTsJ/(the first 2 are the JSFiddle links - you won't have them on your page)
试试这个 - http://jsfiddle.net/TpTsJ/（前两个是 JSFiddle 链接 - 你的页面上不会有它们）
var links = document.getElementsByTagName("link");
for ( var i = 0; i < links.length; i++ ) {
alert( links[i].getAttribute("rel") + ' : ' + links[i].getAttribute("href") );
}?
