我现在正在使用网状拓扑从WebRTC Samples实现源代码以成为3个用户连接。
但是,我的代码不能像我想象的那样工作。我想我停留在调用函数iceCallBack#
(#指的是数字1,2,3),因为输出结果与原始结果相同。它只能连接2个用户。
我不知道如何以正确的方式修复它。
以下是我一直在处理的一些JavaScript代码:
var audio2 = document.querySelector('audio#audio2');
var audio3 = document.querySelector('audio#audio3');
var pc1;
var pc2;
var pc3;
function call() {
callButton.disabled = true;
codecSelector.disabled = true;
trace('Starting call');
var servers = null;
var pcConstraints = {
'optional': []
};
pc1 = new RTCPeerConnection(servers, pcConstraints);
trace('Created local peer connection object pc1');
pc1.onicecandidate = iceCallback1;
pc2 = new RTCPeerConnection(servers, pcConstraints);
trace('Created remote peer connection object pc2');
pc2.onicecandidate = iceCallback2;
pc2.onaddstream = gotRemoteStream;
trace('Requesting local stream');
pc3 = new RTCPeerConnection(servers, pcConstraints);
trace('Created remote peer connection object pc2');
pc3.onicecandidate = iceCallback3;
pc3.onaddstream = gotRemoteStream2;
trace('Requesting local stream');
navigator.mediaDevices.getUserMedia({
audio: true,
video: false
})
.then(gotStream)
.catch(function(e) {
alert('getUserMedia() error: ' + e.name);
});
}
//Description of pc1 creating offer to pc2
function gotDescription1(desc) {
desc.sdp = forceChosenAudioCodec(desc.sdp);
trace('Offer from pc1 \n' + desc.sdp);
pc1.setLocalDescription(desc, function() {
pc2.setRemoteDescription(desc, function() {
pc2.createAnswer(gotDescription2, onCreateSessionDescriptionError);
}, onSetSessionDescriptionError);
}, onSetSessionDescriptionError);
}
//Description of pc1 creating offer to pc3
function gotDescription3(desc) {
desc.sdp = forceChosenAudioCodec(desc.sdp);
trace('Offer from pc1 \n' + desc.sdp);
pc1.setLocalDescription(desc, function() {
pc3.setRemoteDescription(desc, function() {
pc3.createAnswer(gotDescription4, onCreateSessionDescriptionError); //Must edit gotDescription4
}, onSetSessionDescriptionError);
}, onSetSessionDescriptionError);
}
//Creating answer from pc2
function gotDescription2(desc) {
desc.sdp = forceChosenAudioCodec(desc.sdp);
pc2.setLocalDescription(desc, function() {
trace('Answer from pc2 \n' + desc.sdp);
pc1.setRemoteDescription(desc, function() {
}, onSetSessionDescriptionError);
}, onSetSessionDescriptionError);
}
//Creating answer from pc3
function gotDescription4(desc) {
desc.sdp = forceChosenAudioCodec(desc.sdp);
pc3.setLocalDescription(desc, function() {
trace('Answer from pc2 \n' + desc.sdp);
pc1.setRemoteDescription(desc, function() {
}, onSetSessionDescriptionError);
}, onSetSessionDescriptionError);
}
function iceCallback1(event) {
if (event.candidate) {
pc2.addIceCandidate(new RTCIceCandidate(event.candidate),
onAddIceCandidateSuccess, onAddIceCandidateError);
pc3.addIceCandidate(new RTCIceCandidate(event.candidate),
onAddIceCandidateSuccess, onAddIceCandidateError);
trace('Local ICE candidate: \n' + event.candidate.candidate);
}
}
function iceCallback2(event) {
if (event.candidate) {
pc1.addIceCandidate(new RTCIceCandidate(event.candidate),
onAddIceCandidateSuccess, onAddIceCandidateError);
pc3.addIceCandidate(new RTCIceCandidate(event.candidate),
onAddIceCandidateSuccess, onAddIceCandidateError);
trace('Remote ICE candidate: \n ' + event.candidate.candidate);
}
}
function iceCallback3(event) {
if (event.candidate) {
pc1.addIceCandidate(new RTCIceCandidate(event.candidate),
onAddIceCandidateSuccess, onAddIceCandidateError);
pc2.addIceCandidate(new RTCIceCandidate(event.candidate),
onAddIceCandidateSuccess, onAddIceCandidateError);
trace('Remote ICE candidate: \n ' + event.candidate.candidate);
}
}
<div id="audio">
<div>
<div class="label">Local audio:</div><audio id="audio1" autoplay controls muted></audio>
</div>
<div>
<div class="label">Remote audio2:</div><audio id="audio2" autoplay controls></audio>
</div>
<div>
<div class="label">Remote audio3:</div><audio id="audio3" autoplay controls></audio>
</div>
</div>
注意:我是webRTC的新手。我可能会以某种方式愚蠢,请原谅我。
非常感谢你。
答案 0 :(得分:5)
3个用户的网格意味着每个用户设置两个连接,其中一个连接到另外两个用户。在每个客户端,这两个完全不同的RTCPeerConnections,并且您不能在它们之间重用候选,因为每个候选包含专门为媒体分配的端口号以及要发送给它的目标。 / p>
如果您知道如何设置一个连接,则知道如何设置两个连接。只是把事情分开。