i want to download a video from this URL-CodePudding

import urllib
link = "https://www.facebook.com/cofidis/videos/1518507675176091/?__cft__[0]=AZWA0HX9K9N1U6ZGKugETU-gWHDvRPDv0Vg7kmZwd2cR45pt9_Uml9gYz33wD3taI1aZc_w18hmTiUVd7VIXQxQTecK5v76B6r7gUdBxf4V1mAJgbV77A_NMi-07ywYyAQrUFVIT0if2gNQpFMS8Dq91o0w3xwALSDf0WV7_41Vw7EQUMYldqmaU6ZE29VkNXE4&__tn__=+?FH-R"
urllib.request.urlretrieve(link, "video.mp4")

CodePudding user response：

import urllib

link = "https://www.facebook.com/cofidis/videos/1518507675176091/?cft[0]=AZWA0HX9K9N1U6ZGKugETU-gWHDvRPDv0Vg7kmZwd2cR45pt9_Uml9gYz33wD3taI1aZc_w18hmTiUVd7VIXQxQTecK5v76B6r7gUdBxf4V1mAJgbV77A_NMi-07ywYyAQrUFVIT0if2gNQpFMS8Dq91o0w3xwALSDf0WV7_41Vw7EQUMYldqmaU6ZE29VkNXE4&tn=+?FH-R"

opener = urllib.FancyURLopener()

opener.retrieve(link,"testout.flv")

CodePudding user response：

I think this is the simplest technique to achieve your aim.

downloader = urllib.FancyURLopener()
downloader.retrieve(link,"video.mp4")

Or,

file_name = 'video.mp4' 
urllib.retrieve(link, file_name)