2

当我单击一台摄像机时，我的计算机中附有不同摄像机的列表，它显示它在视频标签中的蒸汽，然后在我单击列表中的另一台摄像机后，它在流媒体中显示新的视频标签，但问题是我想显示流媒体第一次显示的同一视频中的视频

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>
4

1 回答 1

2

这是一个完全的猜测，因为我不流式传输视频，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();

对此

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();
}
于 2013-09-20T09:01:44.977 回答