CODEWITHSUNDEEP
htmlvideotagssequenceplaying
Frankie 'Moodurian' Kam
Frankie 'Moodurian' Kam

Reputation: 157

PLaying two videos in sequence in Chrome by using the <video> tag

How do I play two videos in a sequence in the HTML5 video tag?

In Google Chrome, the following code plays only the first intro video.

<html>
<head>
<script src="http://code.jquery.com/jquery-1.4.2.min.js"></script>
<script>

var i = 0;
var sources = ['1.mp4', '2.mp4'];
videoElement.addEventListener('ended', function(){
   videoElement.src = sources[(++i)%sources.length];
   videoElement.load();
   videoElement.play();
}, true);

</script>

</head>
<body>
<video id="videoElement" width="640" height="360" autoplay="autoplay">
    <source src="intro.mp4" type="video/mp4"></source>
</video>

<body>
<html>

Upvotes: 0

Views: 8722

Answers (2)

Juraj.Lorinc
Juraj.Lorinc

Reputation: 533

In case someone came across this question again, here is my solution to similar problem-I needed to play first video once and then second video in a loop. I also have support for .webm, .m4v and .mp4.

This is my JS:

    $(document).ready(function(){
        var vid = document.getElementById("landing-video");
        vid.onplay = function() {
            var source=vid.currentSrc;
            folder = source.match(/(.+)(\/)/);
            ext = source.match(/(\.\w+)$/);
        };

        vid.onended = function() {
          $("#landing-video").attr({
            "src":folder[0]+"video-2"+ext[0],
            "loop":""
          });
        };
    });

And this is my HTML:

    <video autoplay="" muted="" poster="" id="landing-video">
      <source src="my-folder/video-1.webm" type="video/webm">
      <source src="my-folder/video-1.m4v" type="video/x-m4v">
      <source src="my-folder/video-1.mp4" type="video/mp4">
    </video>

This might save someone some time.

Upvotes: 2

ExpExc
ExpExc

Reputation: 3926

Browser should fire error 'videoElement is not defined' with your JavaScript code, you must get video element from DOM instead of using its id directly. Please change your code to

$(document).ready(function() {
    //place code inside jQuery ready event handler 
    //to ensure videoElement is available
    var i = 0;
    var sources = ['1.mp4', '2.mp4'];
    $('#videoElement').bind('ended', function() {
        //'this' is the DOM video element
        this.src = sources[i++ % sources.length];
        this.load();
        this.play();
    });
});

Upvotes: 4

Related Questions