我有这个代码每秒从相机/音频中捕获/记录视频$(function () { var handleSuccess = function(stream) { var player = document.querySelector("#vid-user"); var guestPlayer = document.querySelector("#vid-guest"); player.srcObject = stream; console.log("Starting media recording") var options = {mimeType: 'video/webm'}; var mediaRecorder = new MediaRecorder(stream, options); mediaRecorder.ondataavailable = function(e) { console.log("Data available") if (e.data.size > 0) { // How do I display the captured video to the guestPlayer ? } } mediaRecorder.start(1000); }; navigator.mediaDevices.getUserMedia({ audio: true, video: true }) .then(handleSuccess)})这里的问题是如何将捕获的视频显示给guestPlayer我正在这样做以测试捕获的数据,因为该 webm 块e.data将被上传到服务器。e.data包含_data: Blobsize: 26009type: "video/webm"__proto__: Blobsize: (...)type: (...)slice: ƒ slice()stream: ƒ stream()text: ƒ text()arrayBuffer: ƒ arrayBuffer()constructor: ƒ Blob()Symbol(Symbol.toStringTag): "Blob"get size: ƒ size()get type: ƒ type()__proto__: Object
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桃花长相依
TA贡献1860条经验 获得超8个赞
就这么简单:
video.src = URL.createObjectURL(e.data);
https://developer.mozilla.org/en-US/docs/Web/API/URL/createObjectURL
请务必在使用URL.revokeObjectURL()
完该 blob 后使用。
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