Muh*_*gan 4 javascript iframe youtube-api youtube-javascript-api youtube-iframe-api
这个页面iframe上有一个我想从标签中获取屏幕截图,所以我必须到达标签中的视频标签。videoiframe
当我打开控制台并运行此代码时:
const videoElement = document.getElementsByTagName('iframe')[0]
.contentWindow.document.getElementsByTagName('video')[0];
//Extracting picture from video tag
const canvas = document.createElement('canvas');
canvas.width = videoElement.videoWidth;
canvas.height = videoElement.videoHeight;
canvas.getContext('2d').drawImage(videoElement, 0, 0, canvas.width, canvas.height);
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已抛出此错误:
Uncaught DOMException: Blocked a frame with origin "https://developers.google.com" from accessing a cross-origin frame.
at <anonymous>:1:57
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另外,我检查了这个问题
我的问题是如何从 YouTube Player API 获取屏幕截图?
小智 8
据我所知,无法从 YouTube Player API 截取屏幕截图,因为它基于 iFrame。如果你想在你自己的应用程序中制作它们(不仅仅是浏览器扩展),这个操作将被 CORS 禁止(你得到的异常的原因)。
唯一的解决方法是使用您可以从 YouTube 获得的数据将 YouTube 视频作为源放入视频 HTML 元素中。此代码应该可以方便地获取视频的源网址:
class YoutubeVideo {
constructor(video_id, callback) {
return (async () => {
// You should also redirect those requests
// through your own API that would permit CORS
const response = await fetch(`https://www.youtube.com/get_video_info?video_id=${video_id}`, {
headers: { 'Content-Type' : 'text/plain'}
});
const video_info = await response.text();
let video = this.decodeQueryString(video_info);
if (video.status === 'fail') {
return callback(video);
}
if (video.url_encoded_fmt_stream_map)
video.source = this.decodeStreamMap(video.url_encoded_fmt_stream_map);
return callback(video);
})();
}
decodeQueryString(queryString) {
var key, keyValPair, keyValPairs, r, val, _i, _len;
r = {};
keyValPairs = queryString.split("&");
for (_i = 0, _len = keyValPairs.length; _i < _len; _i++) {
keyValPair = keyValPairs[_i];
key = decodeURIComponent(keyValPair.split("=")[0]);
val = decodeURIComponent(keyValPair.split("=")[1] || "");
r[key] = val;
}
return r;
}
decodeStreamMap(url_encoded_fmt_stream_map) {
var quality, sources, stream, type, urlEncodedStream, _i, _len, _ref;
sources = {};
_ref = url_encoded_fmt_stream_map.split(",");
for (_i = 0, _len = _ref.length; _i < _len; _i++) {
urlEncodedStream = _ref[_i];
stream = this.decodeQueryString(urlEncodedStream);
type = stream.type.split(";")[0];
quality = stream.quality.split(",")[0];
stream.original_url = stream.url;
stream.url = "" + stream.url + "&signature=" + stream.sig;
sources["" + type + " " + quality] = stream;
}
return sources;
}
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传递给构造函数中回调的对象将具有包含所有可用视频类型和质量的源链接的 source 属性,您可以在浏览器的控制台中更好地检查它们。尽管如此,并不是所有的 YouTube 视频都可以这样处理,当您只能获得禁止错误或空源时,我遇到了具有进一步限制的文件。
帮助我找到此解决方案的资源:https : //github.com/endlesshack/youtube-video
基于此解决方案的资源:http: //youtubescreenshot.com/
概念证明基于 expressjs 服务器的简单 Web App:https : //github.com/RinSer/YouCut