我正在尝试通过拼凑示例来学习在线教程。我觉得这应该是播放mp3文件。我使用的是 Chrome 浏览器，它是最新的。我在控制台上没有收到任何错误。我不确定我需要更改或添加什么才能使这项工作正常进行。
<html>
<head>
<script type="text/javascript">
var context;
var sound1Buffer = null;
var url = 'https://dl.dropboxusercontent.com/u/1957768/SrtV2.mp3';
function init(){
try {
window.AudioContext = window.AudioContext || window.webkitAudioContext;
context = new AudioContext();
}
catch(e) {
alert("web Audio api is not supported!");
}
}
window.addEventListener('load', init, false);
function loadDogSound(url){
var request = new XMLHttpRequest();
request.open("GET", url, true);
request.responseType = 'arrayBuffer';
//decode asynchronously
request.onload = function(){
context.decodeAudioData(request.response, function(buffer){
sound1Buffer = buffer;
}, onError);
}
request.send();
}
function playSound(sound1Buffer){
var source = context.createBufferSource();
source.sound1Buffer = sound1Buffer;
source.connect(context.destination);
source.start(0);
}
</script>
</head>
<body>
</body>
</html>
最佳答案
您永远不会调用 loadDogSound。如果你调用它，你会发现你确实得到了一个错误:
Uncaught ReferenceError: onError is not defined
此外，您永远不会调用 playSound。
这是一个工作示例:
<html>
<head>
<script type="text/javascript">
//That one url you wanted.
var url = 'https://dl.dropboxusercontent.com/u/1957768/SrtV2.mp3';
/* --- set up web audio --- */
//create the context
var context = new webkitAudioContext();
//...and the source
var source = context.createBufferSource();
//connect it to the destination so you can hear it.
source.connect(context.destination);
/* --- load up that buffer --- */
//Basic start to ajax! (I say basic, yet i don't know it well.)
var request = new XMLHttpRequest();
//open the request...?
request.open('GET', url, true);
//I don't even know.
request.responseType = 'arraybuffer';
//Once the request has completed... do this
request.onload = function() {
context.decodeAudioData(request.response, function(response) {
/* --- play the sound AFTER we've gotten the buffer loaded --- */
//set the buffer to the response we just received.
source.buffer = response;
//And off we go! .start(0) should play asap.
source.start(0);
}, function () { console.error('The request failed.'); } );
}
//Now that the request has been defined, actually make the request. (send it)
request.send();
</script>
</head>
<body>
</body>
</html>
关于javascript - 基本网络音频 API 不播放声音，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/17310411/