Python: Downloading an image from an URL but getting HTTP Error 403

Question

I'm trying to download this image: https://bu3.mkklcdnbuv1.com/mangakakalot/m2/mother_im_sorry/chapter_5_chapter_5/2.jpg

I have tried: Set Headers = {'User-agent': 'Mozilla/5.0'}

The code i received is still 403.

Can anyone suggest me a way to overcome this?

My Code:

import requests
import shutil

r = requests.get('https://bu3.mkklcdnbuv1.com/mangakakalot/m2/mother_im_sorry/chapter_5_chapter_5/2.jpg',stream=True, headers={'User-agent': 'Mozilla/5.0'})
print (r.status_code)
if r.status_code == 200:
    with open("img.jgp", 'wb') as f:
        r.raw.decode_content = True
        shutil.copyfileobj(r.raw, f)

Oh. I realized if i paste that URL to browser, i can't access it, but i can if i go to mangakakalot and select a random chapter => open image in new tab then i can... — Tran Anh Tuan
– Tran Anh Tuan, Commented Sep 6, 2020 at 11:47

jizhihaoSAMA · Accepted Answer · 2020-09-06 12:28:42Z

2

You need to set the referer header in your code, like:

import requests

with requests.Session() as session:
    resp_2 = session.get("https://bu3.mkklcdnbuv1.com/mangakakalot/m2/mother_im_sorry/chapter_5_chapter_5/2.jpg", headers={"referer":"https://mangakakalot.com/chapter/ro920198/chapter_5"})
    with open("xx.jpg","wb") as f:
        f.write(resp_2.content)

edited Sep 6, 2020 at 12:28

answered Sep 6, 2020 at 12:00

jizhihaoSAMA

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1 Comment

Tran Anh Tuan Over a year ago

Thanks alot! But how can i decode and save the content? I've tried: r.content.decode() but it didn't work...

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