After you have the link, request the link.   try: response = requests.request("POST", urlApi, headers=header, data=payload) except: print ('error') print (response.status_code) if response.status_code == 202: urlVideo = response.text time.sleep(5) # wait 5 sec then the image is availble dc_data = eval(response.text) urlVideo = (dc_data['url']) resp = requests.request("GET", urlVideo, stream=True) fn = 'imagename.jpg' try: with open(fn, 'wb') as f: for chunk in resp: f.write(chunk) f.close()   ... View more

Thank you very much for the quick reply, I will try using selenium but that is too slow for what i thought the API should be doing in a more simple faster way.   Thank you very much ... View more

+1 for "Do you plan to implement cloud storage (microsoft, google, amazon)" We want the ability to send footage to our own cloud account for further processing and archiving. ... View more