HTTP retrival using python3

Question

import os
import urllib.request 


foldername="test_1"
os.makedirs(str(foldername))
tempurl=r"http://img2.wikia.nocookie.net/__cb20130802012237/camp-tv-series/images/thumb/e/e5/Adam.jpg/250px-Adam.jpg"

fullpath1=r"C:\Users\\files"+foldername
urllib.request.urlretrieve(tempurl, fullpath1)

This code is creating the folder but not retriving the file. if I give obsolute path instead of fullpath it works. am I missing anything

Initially I gave a test url. now I am giving an example of a obsolute url

why are you doing str(foldername) when it's already a string? — sansa
– sansa, Commented Nov 28, 2019 at 22:28
Did you add the path separator (e.g. "\\") before the foldername? — Kent Shikama
– Kent Shikama, Commented Nov 28, 2019 at 22:28
Also I suggest using docs.python.org/3.5/library/os.path.html#os.path.join instead of a hard coded path separator. — Kent Shikama
– Kent Shikama, Commented Nov 28, 2019 at 22:29

sansa · Accepted Answer · 2019-12-04 01:22:59Z

1

The problem is that you're doing urllib.request.urlretrieve(tempurl, fullpath1) where fullpath does not specify the actual file being retrieved, change this to urllib.request.urlretrieve(tempurl, os.path.join(fullpath1,"adam.jpg"))

Furthermore, you're using relative path to create the directory then you use absolute path to reference it? Stick to one. In addition to this, you're converting a string to a string? Don't do that, it's already a string. Also, avoid using hardcoded paths. Full solution:

import os
from urllib.request import urlretrieve as get

foldername="test_1"
os.makedirs(foldername, exist_ok=True)
url="http://img2.wikia.nocookie.net/__cb20130802012237/camp-tv-series/images/thumb/e/e5/Adam.jpg/250px-Adam.jpg"
get(url, os.path.join(foldername,"adam.jpg"))

edited Dec 4, 2019 at 1:22

answered Nov 28, 2019 at 22:33

sansa

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Ayoub Benayache · Accepted Answer · 2019-11-28 22:53:18Z

0

try this code it's working, you need to use a correct name

import os
import urllib.request


foldername ="tesd"
os.makedirs(str(foldername)



tempurl =r"https://f1.media.brightcove.com/8/1078702682/1078702682_6004950245001_6004956161001-vs.jpg"

archive_path, headers = urllib.request.urlretrieve(tempurl ,foldername+"/"+str(tempurl.split("/")[-1]), data=None)

edited Nov 28, 2019 at 22:53

answered Nov 28, 2019 at 22:47

Ayoub Benayache

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1 Comment

Ayoub Benayache Over a year ago

the problem is in the name of the downloaded file so i removed the full path and let juste the name of file

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