Which Camera will open getUserMedia API in mobile Device? Front or Rear?

Question

在使用 getUserMedia API 访问桌面摄像头时，它会打开网络摄像头。当然这对视频通信有帮助。但是在移动设备中使用时调用哪个摄像头。前摄像头还是后摄像头？。是否需要代码来选择相机？

Answer 1

There's a solution where the user can select one of the cameras.

Enable rear camera with HTML5

By evaluating sourceInfo.facing from the following code, you can select a camera ('user' or 'environment') (which works on current chrome >= 30): https://simpl.info/getusermedia/sources/

'use strict';

var videoElement = document.querySelector('video');
var audioSelect = document.querySelector('select#audioSource');
var videoSelect = document.querySelector('select#videoSource');

navigator.getUserMedia = navigator.getUserMedia ||
  navigator.webkitGetUserMedia || navigator.mozGetUserMedia;

function gotSources(sourceInfos) {
  for (var i = 0; i !== sourceInfos.length; ++i) {
    var sourceInfo = sourceInfos[i];
    var option = document.createElement('option');
    option.value = sourceInfo.id;
    if (sourceInfo.kind === 'audio') {
      option.text = sourceInfo.label || 'microphone ' + (audioSelect.length + 1);
      audioSelect.appendChild(option);
    } else if (sourceInfo.kind === 'video') {
      option.text = sourceInfo.label || 'camera ' + (videoSelect.length + 1);
      videoSelect.appendChild(option);
    } else {
      console.log('Some other kind of source: ', sourceInfo);
    }
  }
}

if (typeof MediaStreamTrack === 'undefined'){
  alert('This browser does not support MediaStreamTrack.\n\nTry Chrome Canary.');
} else {
  MediaStreamTrack.getSources(gotSources);
}


function successCallback(stream) {
  window.stream = stream; // make stream available to console
  videoElement.src = window.URL.createObjectURL(stream);
  videoElement.play();
}

function errorCallback(error){
  console.log('navigator.getUserMedia error: ', error);
}

function start(){
  if (!!window.stream) {
    videoElement.src = null;
    window.stream.stop();
  }
  var audioSource = audioSelect.value;
  var videoSource = videoSelect.value;
  var constraints = {
    audio: {
      optional: [{sourceId: audioSource}]
    },
    video: {
      optional: [{sourceId: videoSource}]
    }
  };
  navigator.getUserMedia(constraints, successCallback, errorCallback);
}

audioSelect.onchange = start;
videoSelect.onchange = start;

start();

Answer 2

Unfortunately, you cannot (yet?) select this via the code.

• Mozilla Firefox beta: When the user accepts to share camera, he/she also select what camera to share.

• Chrome beta: I have only been able to use the face camera. Could be configurable, but I do not know how…

EDIT: In chrome it's possible to select the back facing camera programmatically. See the next answer by Probot in this thread.

Answer 3

The answer is yes, for Android the default camera is the front( <=> user) one. So I developed thes script below to choose between the front camera and the rear Camera

    //----------------------------------------------------------------------
    //  Here we list all media devices, in order to choose between
    //  the front camera and the rear camera.
    //      videoDevices[0] : Front Camera
    //      videoDevices[1] : Back Camera
    //  I used an array to save the devices ID 
    //  which I get with using devices.forEach()
    //  Then I set the video resolution.
    //----------------------------------------------------------------------
    navigator.mediaDevices.enumerateDevices()
    .then(devices => {
      var videoDevices = [0,0];
      var videoDeviceIndex = 0;
      devices.forEach(function(device) {
        console.log(device.kind + ": " + device.label +
          " id = " + device.deviceId);
        if (device.kind == "videoinput") {  
          videoDevices[videoDeviceIndex++] =  device.deviceId;    
        }
      });

      var constraints =  {width: { min: 1024, ideal: 1280, max: 1920 },
      height: { min: 776, ideal: 720, max: 1080 },
      deviceId: { exact: videoDevices[1]  } 
    };
    return navigator.mediaDevices.getUserMedia({ video: constraints });

  })
    .then(stream => {
      if (window.webkitURL) {
        video.src = window.webkitURL.createObjectURL(stream);
        localMediaStream = stream;
      } else if (video.mozSrcObject !== undefined) {
        video.mozSrcObject = stream;
      } else if (video.srcObject !== undefined) {
        video.srcObject = stream;
      } else {
        video.src = stream;
      }})
    .catch(e => console.error(e));