Вот так работает:
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: 'http://wotbonuscode.tk/getcode.php',
type: 'GET',
success: function(res) {
alert(res.responseText);
}
});
Но возвращает:
|
Код:
|
<html>
<head>
<meta content="HTML Tidy for Java (vers. 26 Sep 2004), see www.w3.org" name="generator"/>
<title/>
</head>
<body>
<p>TESTCODE</p>
</body>
</html> |
А мне нужно просто "TESTCODE", как это сделать?