Phân tích cú pháp thức ăn
(Đừng thực sự khuyên bạn rằng, hãy xem các tùy chọn khác.)
jQuery.getFeed({
url : FEED_URL,
success : function (feed) {
console.log(feed.title);
// do more stuff here
}
});
Với Hỗ trợ XML tích hợp của jQuery
$.get(FEED_URL, function (data) {
$(data).find("entry").each(function () { // or "item" or whatever suits your feed
var el = $(this);
console.log("------------------------");
console.log("title : " + el.find("title").text());
console.log("author : " + el.find("author").text());
console.log("description: " + el.find("description").text());
});
});
$.ajax({
url : document.location.protocol + '//ajax.googleapis.com/ajax/services/feed/load?v=1.0&num=10&callback=?&q=' + encodeURIComponent(FEED_URL),
dataType : 'json',
success : function (data) {
if (data.responseData.feed && data.responseData.feed.entries) {
$.each(data.responseData.feed.entries, function (i, e) {
console.log("------------------------");
console.log("title : " + e.title);
console.log("author : " + e.author);
console.log("description: " + e.description);
});
}
}
});
Nhưng điều đó có nghĩa là bạn tin tưởng vào việc họ đang trực tuyến và có thể truy cập.
Xây dựng nội dung
Khi bạn đã trích xuất thành công thông tin bạn cần từ nguồn cấp dữ liệu, bạn có thể tạo DocumentFragments (có document.createDocumentFragment()chứa các yếu tố (được tạo bằng document.createElement()) bạn sẽ muốn tiêm để hiển thị dữ liệu của mình.
Tiêm nội dung
Chọn thành phần chứa mà bạn muốn trên trang và nối các đoạn tài liệu của bạn vào đó và chỉ cần sử dụng InternalHTML để thay thế hoàn toàn nội dung của nó.
Cái gì đó như:
$('#rss-viewer').append(aDocumentFragmentEntry);
hoặc là:
$('#rss-viewer')[0].innerHTML = aDocumentFragmentOfAllEntries.innerHTML;
Kiểm tra dữ liệu
Sử dụng nguồn cấp dữ liệu của câu hỏi này , mà khi viết bài này đưa ra:
<?xml version="1.0" encoding="utf-8"?>
<feed xmlns="http://www.w3.org/2005/Atom" xmlns:creativeCommons="http://backend.userland.com/creativeCommonsRssModule" xmlns:re="http://purl.org/atompub/rank/1.0">
<title type="text">How to parse a RSS feed using javascript? - Stack Overflow</title>
<link rel="self" href="https://stackoverflow.com/feeds/question/10943544" type="application/atom+xml" />
<link rel="hub" href="http://pubsubhubbub.appspot.com/" />
<link rel="alternate" href="https://stackoverflow.com/q/10943544" type="text/html" />
<subtitle>most recent 30 from stackoverflow.com</subtitle>
<updated>2012-06-08T06:36:47Z</updated>
<id>https://stackoverflow.com/feeds/question/10943544</id>
<creativeCommons:license>http://www.creativecommons.org/licenses/by-sa/3.0/rdf</creativeCommons:license>
<entry>
<id>https://stackoverflow.com/q/10943544</id>
<re:rank scheme="http://stackoverflow.com">2</re:rank>
<title type="text">How to parse a RSS feed using javascript?</title>
<category scheme="https://stackoverflow.com/feeds/question/10943544/tags" term="javascript"/><category scheme="https://stackoverflow.com/feeds/question/10943544/tags" term="html5"/><category scheme="https://stackoverflow.com/feeds/question/10943544/tags" term="jquery-mobile"/>
<author>
<name>Thiru</name>
<uri>https://stackoverflow.com/users/1126255</uri>
</author>
<link rel="alternate" href="/programming/10943544/how-to-parse-a-rss-feed-using-javascript" />
<published>2012-06-08T05:34:16Z</published>
<updated>2012-06-08T06:35:22Z</updated>
<summary type="html">
<p>I need to parse the RSS-Feed(XML version2.0) using XML and I want to display the parsed detail in HTML page, I tried in many ways. But its not working. My system is running under proxy, since I am new to this field, I don't know whether it is possible or not. If any one knows please help me on this. Thanks in advance.</p>
</summary>
</entry>
<entry>
<id>https://stackoverflow.com/questions/10943544/-/10943610#10943610</id>
<re:rank scheme="http://stackoverflow.com">1</re:rank>
<title type="text">Answer by haylem for How to parse a RSS feed using javascript?</title>
<author>
<name>haylem</name>
<uri>https://stackoverflow.com/users/453590</uri>
</author>
<link rel="alternate" href="/programming/10943544/how-to-parse-a-rss-feed-using-javascript/10943610#10943610" />
<published>2012-06-08T05:43:24Z</published>
<updated>2012-06-08T06:35:22Z</updated>
<summary type="html"><h1>Parsing the Feed</h1>
<h3>With jQuery's jFeed</h3>
<p>Try this, with the <a href="http://plugins.jquery.com/project/jFeed" rel="nofollow">jFeed</a> <a href="http://www.jquery.com/" rel="nofollow">jQuery</a> plug-in</p>
<pre><code>jQuery.getFeed({
url : FEED_URL,
success : function (feed) {
console.log(feed.title);
// do more stuff here
}
});
</code></pre>
<h3>With jQuery's Built-in XML Support</h3>
<pre><code>$.get(FEED_URL, function (data) {
$(data).find("entry").each(function () { // or "item" or whatever suits your feed
var el = $(this);
console.log("------------------------");
console.log("title : " + el.find("title").text());
console.log("author : " + el.find("author").text());
console.log("description: " + el.find("description").text());
});
});
</code></pre>
<h3>With jQuery and the Google AJAX APIs</h3>
<p>Otherwise, <a href="https://developers.google.com/feed/" rel="nofollow">Google's AJAX Feed API</a> allows you to get the feed as a JSON object:</p>
<pre><code>$.ajax({
url : document.location.protocol + '//ajax.googleapis.com/ajax/services/feed/load?v=1.0&amp;num=10&amp;callback=?&amp;q=' + encodeURIComponent(FEED_URL),
dataType : 'json',
success : function (data) {
if (data.responseData.feed &amp;&amp; data.responseData.feed.entries) {
$.each(data.responseData.feed.entries, function (i, e) {
console.log("------------------------");
console.log("title : " + e.title);
console.log("author : " + e.author);
console.log("description: " + e.description);
});
}
}
});
</code></pre>
<p>But that means you're relient on them being online and reachable.</p>
<hr>
<h1>Building Content</h1>
<p>Once you've successfully extracted the information you need from the feed, you need to create document fragments containing the elements you'll want to inject to display your data.</p>
<hr>
<h1>Injecting the content</h1>
<p>Select the container element that you want on the page and append your document fragments to it, and simply use innerHTML to replace its content entirely.</p>
</summary>
</entry></feed>
Thi hành
Sử dụng Hỗ trợ XML tích hợp của jQuery
Kêu gọi:
$.get('https://stackoverflow.com/feeds/question/10943544', function (data) {
$(data).find("entry").each(function () { // or "item" or whatever suits your feed
var el = $(this);
console.log("------------------------");
console.log("title : " + el.find("title").text());
console.log("author : " + el.find("author").text());
console.log("description: " + el.find("description").text());
});
});
In ra:
------------------------
title : How to parse a RSS feed using javascript?
author :
Thiru
https://stackoverflow.com/users/1126255
description:
------------------------
title : Answer by haylem for How to parse a RSS feed using javascript?
author :
haylem
https://stackoverflow.com/users/453590
description:
Sử dụng jQuery và API AJAX của Google
Kêu gọi:
$.ajax({
url : document.location.protocol + '//ajax.googleapis.com/ajax/services/feed/load?v=1.0&num=10&callback=?&q=' + encodeURIComponent('https://stackoverflow.com/feeds/question/10943544'),
dataType : 'json',
success : function (data) {
if (data.responseData.feed && data.responseData.feed.entries) {
$.each(data.responseData.feed.entries, function (i, e) {
console.log("------------------------");
console.log("title : " + e.title);
console.log("author : " + e.author);
console.log("description: " + e.description);
});
}
}
});
In ra:
------------------------
title : How to parse a RSS feed using javascript?
author : Thiru
description: undefined
------------------------
title : Answer by haylem for How to parse a RSS feed using javascript?
author : haylem
description: undefined