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