如何通过输入字段发送捕获的网络摄像头图像并保存到服务器

问题描述 投票:0回答:1

我正在一个项目中,用户可以使用其网络摄像头来更改其个人资料照片。我通过网络摄像头成功捕获了图像,但是无法将捕获的图像附加到我的图像中形式

这是我的HTML代码

<div class="form-group">
<label for="exampleInputFile">Profile Picture</label>
    <div class="input-group">
        <div class="custom-file">
            <input type="file" id="photo" class="custom-file-input" name = "profile_pic" accept="image/png,image/jpg,image/jpeg">
            <label class="custom-file-label" for="exampleInputFile">Choose Image</label>
        </div>
    </div>
</div>


<div class="camera">
    <video id="video">Video stream not available.</video>
    <button type="button" id="startbutton" class="btn btn-success" data-toggle="modal" data- 
    target="#modal-success">Take photo</button>
</div>

这是我的js代码,我遵循此tutorial寻求帮助

<script>
  (function() {
      // The width and height of the captured photo. We will set the
      // width to the value defined here, but the height will be
      // calculated based on the aspect ratio of the input stream.

      var width = 770;    // We will scale the photo width to this
      var height = 900;   // This will be computed based on the input stream

      // |streaming| indicates whether or not we're currently streaming
      // video from the camera. Obviously, we start at false.

      var streaming = false;

      // The various HTML elements we need to configure or control. These
      // will be set by the startup() function.

      var video = null;
      var canvas = null;
      var photo = null;
      var startbutton = null;

      function startup() {
        video = document.getElementById('video');
        canvas = document.getElementById('canvas');
        photo = document.getElementById('photo');
        startbutton = document.getElementById('startbutton');

        navigator.mediaDevices.getUserMedia({video: true, audio: false})
        .then(function(stream) {
          video.srcObject = stream;
          video.play();
        })
        .catch(function(err) {
          console.log("An error occurred: " + err);
        });

        video.addEventListener('canplay', function(ev){
          if (!streaming) {
            height = video.videoHeight / (video.videoWidth/width);

            // Firefox currently has a bug where the height can't be read from
            // the video, so we will make assumptions if this happens.

            if (isNaN(height)) {
              height = width / (4/3);
            }

            video.setAttribute('width', 340);
            video.setAttribute('height', 300);
            canvas.setAttribute('width', width);
            canvas.setAttribute('height', height);
            streaming = true;
          }
        }, false);

        startbutton.addEventListener('click', function(ev){
          takepicture();
          ev.preventDefault();
        }, false);

        clearphoto();
      }

      // Fill the photo with an indication that none has been
      // captured.

      function clearphoto() {
        var context = canvas.getContext('2d');
        context.fillStyle = "#AAA";
        context.fillRect(0, 0, canvas.width, canvas.height);

        var data = canvas.toDataURL('image/png');
        photo.setAttribute('src', data);
      }

      // Capture a photo by fetching the current contents of the video
      // and drawing it into a canvas, then converting that to a PNG
      // format data URL. By drawing it on an offscreen canvas and then
      // drawing that to the screen, we can change its size and/or apply
      // other changes before drawing it.

      function takepicture() {
        var context = canvas.getContext('2d');
        if (width && height) {
          canvas.width = width;
          canvas.height = height;
          context.drawImage(video, 0, 0, width, height);

          var data = canvas.toDataURL('image/ ');
          photo.setAttribute('src', data);
        } else {
          clearphoto();
        }
      }

      // Set up our event listener to run the startup process
      // once loading is complete.
      window.addEventListener('load', startup, false);
    })(); 



</script>

现在如何将捕获的图像放入<input type="file">字段,用于上传到服务器

注意:我不想为此使用ajax。

javascript django image server webcam
1个回答
0
投票

1-将base64图像作为字符串发送,然后将其转换为服务器端的文件。

2-将Base64字符串转换为Blob以将其作为文件上传到服务器。

我解释第二种选择:

首先,使用此函数将Base64字符串转换为Blob:

function b64toBlob(b64Data, contentType, sliceSize) {
    contentType = contentType || '';
    sliceSize = sliceSize || 512;

    var byteCharacters = atob(b64Data); // window.atob(b64Data)
    var byteArrays = [];

    for (var offset = 0; offset < byteCharacters.length; offset += sliceSize) {
        var slice = byteCharacters.slice(offset, offset + sliceSize);

        var byteNumbers = new Array(slice.length);
        for (var i = 0; i < slice.length; i++) {
            byteNumbers[i] = slice.charCodeAt(i);
        }

        var byteArray = new Uint8Array(byteNumbers);

        byteArrays.push(byteArray);
    }

    var blob = new Blob(byteArrays, {type: contentType});
    return blob;
}

现在,创建您的表单(您可以使用CSS将其隐藏):

<form id="myAwesomeForm" method="post">
    <input type="text" id="filename" name="filename" /> <!-- Filename -->
</form>

然后,使用FormData将图像附加到表单中:

var form = document.getElementById("myAwesomeForm");

var ImageURL = photo; // 'photo' is your base64 image
// Split the base64 string in data and contentType
var block = ImageURL.split(";");
// Get the content type of the image
var contentType = block[0].split(":")[1];// In this case "image/gif"
// get the real base64 content of the file
var realData = block[1].split(",")[1];

// Convert it to a blob to upload
var blob = b64toBlob(realData, contentType);

// Create a FormData and append the file with "image" as parameter name
var formDataToUpload = new FormData(form);
formDataToUpload.append("image", blob);

最后,您可以使用任何方法发送formData,例如:

var request = new XMLHttpRequest();
request.open("POST", "SERVER-URL");
request.send(formDataToUpload);

我希望这可以对您有所帮助;)

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