我正在使用 Django 并尝试使用 XMLHttpRequest 将网络摄像头数据发送到后台 (view.py) 以处理每个帧。我点击此链接,大多数人都为我的问题提出了类似的方法。$(document).ready(function(){ $('#trigger_button').click(function(){ navigator.mediaDevices.getUserMedia(constraints) .then(stream => { document.getElementById("myVideo").srcObject = stream; }) .catch(err => { alert('navigator.getUserMedia error: ', err) }); drawCanvas.width = v.videoWidth; drawCanvas.height = v.videoHeight; imageCanvas.width = uploadWidth; imageCanvas.height = uploadWidth * (v.videoHeight / v.videoWidth); drawCtx.lineWidth = 4; drawCtx.strokeStyle = "cyan"; drawCtx.font = "20px Verdana"; drawCtx.fillStyle = "cyan"; imageCanvas.getContext("2d").drawImage(v, 0, 0, v.videoWidth, v.videoHeight, 0, 0, uploadWidth, uploadWidth * (v.videoHeight / v.videoWidth)); imageCanvas.toBlob(postFile, 'image/jpeg'); });});function postFile(file) { var formdata = new FormData(); formdata.append("image", file); formdata.append("threshold", scoreThreshold); var xhr = new XMLHttpRequest(); xhr.open('POST', apiServer, true); xhr.onload = function () { if (this.status === 200) { var objects = JSON.parse(this.response); drawBoxes(objects); imageCtx.drawImage(v, 0, 0, v.videoWidth, v.videoHeight, 0, 0, uploadWidth, uploadWidth * (v.videoHeight / v.videoWidth)); imageCanvas.toBlob(postFile, 'image/jpeg'); } else { alert(this.status); } }; xhr.send(formdata);}然后,我尝试访问 view.py 中请求中的数据,如下所示:def control4(request): print(request.POST.get('image')) print(request.POST.get('threshold')) return render(request, 'local.html')但是,虽然 request.Post.get('threshold') 返回一个值,但 request.POST.get('image') 返回 None 。另外,该方法重复3次后就停止发送反馈。我的意思是,control4 函数打印 3 次(我认为它应该一直工作到相机关闭为止)。谁能知道问题出在哪里吗?
1 回答
慕沐林林
TA贡献2016条经验 获得超9个赞
你必须寻找request.FILES图像
def control4(request):
print(request.FILES['image'])
print(request.POST.get('threshold'))
return render(request, 'local.html')
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