【发布时间】:2018-01-16 10:19:27
【问题描述】:
如果跳过一些块,视频元素将无法播放。
HTML:
<div>
<video id="vdo0" autoplay muted width="300"></video>
<video id="vdo1" autoplay width="300"></video>
</div>
<div>
<button id="btnStart">Start</button>
</div>
Javascript:
var vdo0 = document.getElementById('vdo0');
var vdo1 = document.getElementById('vdo1');
var ms = new MediaSource();
var sourceBuffer = null;
var chunks = [];
vdo1.src = URL.createObjectURL(ms);
var i = 0;
var recv = function(){
var chunk = chunks.shift();
if(chunk){
i++;
if(i < 5 || i > 10){ // HERE, skipped some chunks, vdo1 will cannot play
sourceBuffer.appendBuffer(chunk);
}
recv();
}else{
setTimeout(recv, 20);
}
};
ms.addEventListener('sourceopen', function () {
sourceBuffer = ms.addSourceBuffer('video/webm;codecs=vp8');
recv();
}, false);
document.getElementById('btnStart').addEventListener('click', function(){
navigator.mediaDevices.getUserMedia({
video: true,
audio: true
}).then(function (s) {
vdo0.srcObject = s;
vdo0.oncanplay = function(){
var mediaRecorder = new MediaRecorder(s, 'video/webm;codecs=vp8');
mediaRecorder.ondataavailable = function (e) {
if (e.data && e.data.size > 0) {
var reader = new FileReader();
reader.addEventListener("loadend", function () {
var arr = new Uint8Array(reader.result);
chunks.push(arr);
});
reader.readAsArrayBuffer(e.data);
}
};
mediaRecorder.start(20);
};
}).catch(function (err) {
A.log('ERROR: ', err);
});
}, false);
我想做, 当跳过一些块时,vdo1 仍然可以播放。 MediaSource 如何接收任何 blob?
如果不将第一个 recorder-blob 推送到块,vdo1 也无法播放。为什么?
怎么办?
谢谢。
【问题讨论】:
标签: javascript html5-video mediarecorder media-source