Google Glass镜像API上的视频路线图是什么? API是否允许将视频流式传输到设备或从设备流式传输,如玻璃演示视频http://www.youtube.com/watch?v=v1uyQZNg2vE?
所示答案 0 :(得分:8)
镜像API没有已发布的路线图。我们的开发人员预览的部分动机是弄明白。
首先,为了澄清,该视频中显示的视频流是Google+环聊。这是Glass内置的功能。
更新:Glass现在支持视频流。您可以找到完整的文档here。
要添加视频流,请将包含视频网址的多部分POST作为其中一个部分,如下所示:
POST /upload/mirror/v1/timeline HTTP/1.1
Host: www.googleapis.com
Authorization: Bearer {auth token}
Content-Type: multipart/related; boundary="mymultipartboundary"
Content-Length: {length}
--mymultipartboundary
Content-Type: application/json; charset=UTF-8
{ "text": "Skateboarding kittens" }
--mymultipartboundary
Content-Type: video/vnd.google-glass.stream-url
http://example.com/path/to/kittens.mp4
--mymultipartboundary--
答案 1 :(得分:1)
Youtube视频流是可能的。我使用“YoutubeExtractor”命名空间在C#.net中完成了它。从您的视频管解析视频(.mp4)网址并将其流式传输。这是代码。它对我来说很好。复制网址时,获取单击共享后可用的管连接
private static String youtubevideoStream(MainController controller)
{
string link = "http://youtu.be/9uYKISlL7Vg";
IEnumerable<VideoInfo> videoInfos = DownloadUrlResolver.GetDownloadUrls(link);
VideoInfo video = videoInfos.First(info => info.VideoType == VideoType.Mp4 && info.Resolution == 360);
String vLink = video.DownloadUrl;
TimelineItem videocard= new TimelineItem()
{
Text = "Menu Card",
BundleId = "666",
Notification = new NotificationConfig() { Level = "DEFAULT" },
MenuItems = new List<MenuItem>()
{
new MenuItem() {Action = "DELETE"},
}
};
String mediaLink = vLink;
if (!String.IsNullOrEmpty(mediaLink))
{
Stream stream = null;
if (mediaLink.StartsWith("/"))
{
stream = new StreamReader(controller.Server.MapPath(mediaLink)).BaseStream;
}
else
{
HttpWebRequest request = WebRequest.Create(mediaLink) as HttpWebRequest;
request.UseDefaultCredentials = false;
HttpWebResponse response = request.GetResponse() as HttpWebResponse;
byte[] b = null;
using (Stream streamFromWeb = response.GetResponseStream())
using (MemoryStream ms = new MemoryStream())
{
int count = 0;
do
{
byte[] buf = new byte[1024];
count = streamFromWeb.Read(buf, 0, 1024);
ms.Write(buf, 0, count);
} while (streamFromWeb.CanRead && count > 0);
b = ms.ToArray();
stream = new MemoryStream(b);
}
}
controller.Service.Timeline.Insert(videocard, stream, "video/mp4").Upload();
}
else
{
controller.Service.Timeline.Insert(videocard).Fetch();
}