Ali*_*Ali 16 javascript youtube jquery youtube-api
<script type= "text/javascript">
var url = "http://gdata.youtube.com/feeds/api/videos/VA770wpLX-Q?v=2&alt=json-in-script&callback=";
var title;
var description;
var viewcount;
var views;
var author;
$.getJSON(url,
function(data){
title = data.entry.title.$t;
description = data.entry.media$group.media$description.$t;
viewcount = data.entry.yt$statistics.viewCount;
views = numberFormat (viewcount);
author = data.entry.author[0].name.$t;
listInfo (title,description,author,views);
});
</script>
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这就是我的代码从单个视频获取信息,在收到信息后,它调用此函数来显示它:
<script type="text/javascript">
function listInfo (title,description,author,views) {
var html = ['<dl>'];
html.push('<dt>','<span class="titleStyle">', title,'</span><span class="descriptionStyle">',description, '</span><span class="authorStyle">',author,'</span><span class="viewsStyle">',' Views:',views,'</span></dt>');
html.push('</dl>');
document.getElementById("agenda").innerHTML = html.join("");
}
function numberFormat(nStr,prefix){
var prefix = prefix || '';
nStr += '';
x = nStr.split('.');
x1 = x[0];
x2 = x.length > 1 ? '.' + x[1] : '';
var rgx = /(\d+)(\d{3})/;
while (rgx.test(x1))
x1 = x1.replace(rgx, '$1' + ',' + '$2');
return prefix + x1 + x2;
}
</script>
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之后,它将定义List放入一个div,我在一个表中
<table width="485"><tr><td><div id="agenda"></div></td></tr></table>
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所有这些信息都在身体中找到,我似乎无法让它工作,我已经尝试了一个星期了,我找不到任何方法让它工作
小智 15
由于youtube API不允许在单个请求中返回超过50条评论,因此您需要添加名为"start-index"的网址参数,该参数告诉您管理系统您希望从此处获取评论.以下是一个例子.我已经做到了这一点,只要响应JSON返回50条评论,它就会在接下来的50条评论中再次调用该函数.
<html>
<head>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.5.1.min.js"></script>
<script type= "text/javascript">
function getYouTubeInfo() {
$.ajax({
url: "http://gdata.youtube.com/feeds/api/videos/<?php echo $_GET['v']; ?>?v=2&alt=json",
dataType: "jsonp",
success: function (data) { parseresults(data); }
});
}
function parseresults(data) {
var title = data.entry.title.$t;
var description = data.entry.media$group.media$description.$t;
var viewcount = data.entry.yt$statistics.viewCount;
var author = data.entry.author[0].name.$t;
$('#title').html(title);
$('#description').html('<b>Description</b>: ' + description);
$('#extrainfo').html('<b>Author</b>: ' + author + '<br/><b>Views</b>: ' + viewcount);
getComments(data.entry.gd$comments.gd$feedLink.href + '&max-results=50&alt=json', 1);
}
function getComments(commentsURL, startIndex) {
$.ajax({
url: commentsURL + '&start-index=' + startIndex,
dataType: "jsonp",
success: function (data) {
$.each(data.feed.entry, function(key, val) {
$('#comments').append('<br/>Author: ' + val.author[0].name.$t + ', Comment: ' + val.content.$t);
});
if ($(data.feed.entry).size() == 50) { getComments(commentsURL, startIndex + 50); }
}
});
}
$(document).ready(function () {
getYouTubeInfo();
});
</script>
<title>YouTube</title>
</head>
<body bgcolor="grey">
<div align="center">
<br/><br/>
<div id="title" style="color: #dddddd">Could not find a title</div><br/>
<iframe title="Youtube Video Player" width="640" height="390" src="http://www.youtube.com/embed/<?php echo $_GET['v']; ?>?fs=1&autoplay=1&loop=0" frameborder="0" allowfullscreen style="border: 1px solid black"></iframe>
<br/><br/>
<div id="description" style="width:400px; background-color: #dddddd; font-size:10px; text-align:left;">Could not find a description</div>
<div id="extrainfo" style="width:400px; background-color: #dddddd; font-size:10px; text-align:left;">Could not find extra information</div>
<div id="comments" style="width:400px; background-color: #dddddd; font-size:10px; text-align:left;">Could not find comments</div>
</div>
</body>
</html>
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如果您有任何其他问题或者您遇到此代码,请不要犹豫再次询问:-)
祝你好运,汤姆
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