使用jquery更改源代码后重新加载video.js播放器

Question

我设法创建一个脚本,我可以在不重新加载整个页面的情况下更改视频源,但问题是在加载新源之后,播放器不是,我只得到一个黑盒子.

HTML:

<link href="//vjs.zencdn.net/4.12/video-js.css" rel="stylesheet">
<script src="//vjs.zencdn.net/4.12/video.js"></script>
<video id="example_video_1" class="video-js vjs-default-skin vjs-big-play-centered" controls autoplay preload="auto" width="850" height="400" data-setup='{"example_option":true}'>
  <source id="videoMP4" src="http://video-js.zencoder.com/oceans-clip.mp4" type='video/mp4' />
</video>
<div class="menu1">menu1</div>
<div class="menu2">menu2</div>
<div class="menu3">menu3</div>
<div class="menu4">menu4</div>

JavaScript的:

$(".menu1").click(function() {
  document.getElementById("example_video_1").innerHTML = "<source src=\"http:\/\/video-js.zencoder.com\/oceans-clip.mp4\" type='video\/mp4' \/>";
});
$(".menu2").click(function() {
  document.getElementById("example_video_1").innerHTML = "<source src=\"http:\/\/techslides.com\/demos\/sample-videos\/small.mp4\" type='video\/mp4' \/>";
});
$(".menu3").click(function() {
  document.getElementById("example_video_1").innerHTML = "<source src=\"http:\/\/video-js.zencoder.com\/oceans-clip.mp4\" type='video\/mp4' \/>";
});
$(".menu4").click(function() {
  document.getElementById("example_video_1").innerHTML = "<source src=\"http:\/\/techslides.com\/demos\/sample-videos\/small.mp4\" type='video\/mp4' \/>";
});

完整代码和示例:http://codepen.io/BeBeINC/pen/PqydVM

Answer 1

使用该.load()功能.你应该暂停它以防止音频继续播放.

var video = document.getElementById('example_video');
var source = document.getElementById('videoMP4');
$("ELEMENT").click(function() {
    video.pause()
    source.setAttribute('src', 'NEW MP4');
    video.load();
    video.play();
});