Dear all,
Thank you for the help. I managed to setup StreamingHttpResponse to read
the images from the media folder. However I believe that there is a syntax
error with html code <img>. It doesn't display the image. Any help? I am
attaching the code below.Thanks in advance!
def stream_response_generator():
yield "<html><body>\n"
for x in range(1, 10):
rval, frame = settings.CAP.read()
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
cv2.imwrite('media/image.jpeg', gray,
[int(cv2.IMWRITE_JPEG_QUALITY), 90])
time.sleep(.1)
x={"src": "%s/image.jpeg" % (settings.MEDIA_URL), "alt": "Video
Frame"}
yield "<img %s/>" % x
yield "</body></html>\n"
def camera(request):
resp = StreamingHttpResponse(stream_response_generator())
return resp
On Friday, November 15, 2013 11:15:21 AM UTC-5, Alex Karargyris wrote:
>
> I have this simple app that I opens webcam and processes the frames using
> OpenCV. In my views.py I have the following code:
>
> def processImage():
>
> while True:
>
> rval, frame = settings.CAP.read() //Read frames from webcam
> gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY) //Make
> frame grayscale
> pil_img = Image.fromarray(gray)
>
> response = HttpResponse(mimetype="image/png")
> pil_img.save(response, "PNG")
>
> return response
>
> Although there is a while loop the function runs only once because it is
> broken by "return response". How could I change the code to make it run
> constantly?
>
>
>
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