Question

我有一个不同的摄像机附加列表在我的计算机上，当我点击一个摄像机它显示我在视频标签中蒸，然后我点击列表中的另一个视频摄像头它显示我在流媒体中的新视频标签但问题是我想要在第一次显示的同一视频中显示视频流

我的代码：

<!DOCTYPE html>
  <head>
      <meta charset="utf-8">
      <meta http-equiv="X-UA-Compatible" content="IE=edge">
      <title>Video Camera List</title>

      <script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.3/jquery.min.js" ></script>

      <style type="text/css" media="screen">
        video {
          border:1px solid gray;
        }
      </style>
  </head>
  <body>
    <script>
      if (!MediaStreamTrack) document.body.innerHTML = '<h1>Incompatible Browser Detected. Try <strong style="color:red;">Chrome Canary</strong> instead.</h1>';

      var videoSources = [];

      MediaStreamTrack.getSources(function(media_sources) {
        console.log(media_sources);
    //  alert('media_sources : '+media_sources);
        media_sources.forEach(function(media_source){
          if (media_source.kind === 'video') {
            videoSources.push(media_source);
          }
        });

        getMediaSource(videoSources);
      });

      var get_and_show_media = function(id) {
        var constraints = {};
        constraints.video = {
          optional: [{ sourceId: id}]
        };

        navigator.webkitGetUserMedia(constraints, function(stream) {
          console.log('webkitGetUserMedia');
          console.log(constraints);
          console.log(stream);

          var mediaElement = document.createElement('video');
          mediaElement.src = window.URL.createObjectURL(stream);
          document.body.appendChild(mediaElement);
          mediaElement.controls = true;
          mediaElement.play();

        }, function (e) 
        {
       //   alert('Hii');  
          document.body.appendChild(document.createElement('hr'));
          var strong = document.createElement('strong');
          strong.innerHTML = JSON.stringify(e);
          alert('strong.innerHTML : '+strong.innerHTML);
          document.body.appendChild(strong);
        });
      };

      var getMediaSource = function(media) {
        console.log(media);
        media.forEach(function(media_source) {
          if (!media_source) return;

          if (media_source.kind === 'video') 
          {
            // add buttons for each media item
            var button = $('<input/>', {id: media_source.id, value:media_source.id, type:'submit'});
            $("body").append(button);
            // show video on click
            $(document).on("click", "#"+media_source.id, function(e){
              console.log(e);
              console.log(media_source.id);
              get_and_show_media(media_source.id);
            });
          }
        });
      }
    </script>
  </body>
</html>

Answer 1

这是一个总猜测，因为我不流视频，但这似乎是合理的jQuery，所以改变

var mediaElement = document.createElement('video');
mediaElement.src = window.URL.createObjectURL(stream);
document.body.appendChild(mediaElement);
mediaElement.controls = true;
mediaElement.play();

到这个

var mediaElement = $('video');
if(mediaElement.length ==0) {
    mediaElement = document.createElement('video');
    mediaElement.controls = true;
    mediaElement.autoplay = true;
    mediaElement.src = window.URL.createObjectURL(stream);
    document.body.appendChild(mediaElement);
} else {
    mediaElement.attr('src', window.URL.createObjectURL(stream));
    mediaElement.attr('autoplay', true);
    mediaElement.load();
}

如何在同一视频标签中流式传输不同的摄像机

1 个答案: