javascript – 如何在Web音频API中反向播放,同时保留正向版本?
作者:互联网
据我所知,Google Chrome(v21)中的Web Audio API不支持以下内容:
source.playbackRate.value = -1;
我非常希望!我会耐心等待,但与此同时,this post有一个不错的替代解决方案.我已经采用它试图反转我自己从loaded bufferlist加载的音频样本,希望我可以同时拥有每个加载缓冲区的正向和反向版本,如下所示:
function finishedLoading(bufferList) {
for (var it = 0; it < this.urlList.length; ++it) {
storedBuffer[it] = bufferList[it]; // assign bufferList to globals
storedBufferR[it] = bufferList[it];
// attempt to reverse storedBufferR only ...
Array.prototype.reverse.call( storedBufferR[it].getChannelData(0) );
Array.prototype.reverse.call( storedBufferR[it].getChannelData(1) );
}
}
上面的函数确实会反转回放,但它会反转’storedbufferR’和’storedbuffer’,这样所有的缓冲区都会反转!
所以这就是我迷路的地方……我明白Array.protoype会影响所有数组,所以我想我可以看到’storedBuffer’对’storedBuffer’的反向方法是如何影响的.编辑:反向方法只影响有问题的数组
但是有可能重写上面的函数以确保array.protoype只影响我想要反转的存储缓冲区吗?如果没有,还有另一种方法可以存储正向和反向版本吗?
解决方法:
关于共享相同引用的两个AudioBuffers的评论是在正确的轨道上,但解决方案不是创建一个新的AudioContext;相反,你只需要创建一个新的AudioBuffer并克隆包含实际音频样本的底层Float32Arrays.
例如,此函数将克隆AudioBuffer:
function cloneAudioBuffer(audioBuffer){
var channels = [],
numChannels = audioBuffer.numberOfChannels;
//clone the underlying Float32Arrays
for (var i = 0; i < numChannels; i++){
channels[i] = new Float32Array(audioBuffer.getChannelData(i));
}
//create the new AudioBuffer (assuming AudioContext variable is in scope)
var newBuffer = context.createBuffer(
audioBuffer.numberOfChannels,
audioBuffer.length,
audioBuffer.sampleRate
);
//copy the cloned arrays to the new AudioBuffer
for (var i = 0; i < numChannels; i++){
newBuffer.getChannelData(i).set(channels[i]);
}
return newBuffer;
}
然后,修改上面的……
function finishedLoading(bufferList) {
for (var it = 0; it < this.urlList.length; ++it) {
storedBuffer[it] = bufferList[it]; // assign bufferList to globals
storedBufferR[it] = cloneAudioBuffer(bufferList[it]);
// attempt to reverse storedBufferR only ...
Array.prototype.reverse.call( storedBufferR[it].getChannelData(0) );
Array.prototype.reverse.call( storedBufferR[it].getChannelData(1) );
}
}
标签:javascript,html5,google-chrome,web-audio,audiobuffer 来源: https://codeday.me/bug/20190901/1785659.html