从URL使用Python下载没有提供主机错误

Question

我正在尝试制作下载漫画的应用程序，但是每当我尝试下载图像时，它都说没有提供主机。

我真的搜索了，什么也没有。

这是代码：

import requests,bs4
url='https://www.marvel.com/comics/issue/71314/edge_of_spider-geddon_2018_1'
res=requests.get(url,stream=True)
res.raise_for_status()
soup=bs4.BeautifulSoup(res.text)
elem=soup.select('div[class="row-item-image"] img')#.viewer-cnt .row .col-xs-12 #ppp img')
#print(elem)
comicurl='https:'+elem[0].get('src')
res=requests.get(comicurl,stream=True,allow_redirects=True)
res.raise_for_status()
with open(comicurl[comicurl.rfind('/')+1:],'wb') as i:
    for chunk in res.iter_content(100000):
        i.write(chunk)

我希望它可以下载图像，但是它给了我这个错误：

Traceback (most recent call last):
  File "C:\Users\Islam\AppData\Local\Programs\Python\Python36\comicdownloader.py", line 10, in <module>
    res=requests.get(comicurl,stream=True,allow_redirects=True)
  File "C:\Users\Islam\AppData\Local\Programs\Python\Python36\lib\site-packages\requests\api.py", line 75, in get
    return request('get', url, params=params, **kwargs)
  File "C:\Users\Islam\AppData\Local\Programs\Python\Python36\lib\site-packages\requests\api.py", line 60, in request
    return session.request(method=method, url=url, **kwargs)
  File "C:\Users\Islam\AppData\Local\Programs\Python\Python36\lib\site-packages\requests\sessions.py", line 519, in request
    prep = self.prepare_request(req)
  File "C:\Users\Islam\AppData\Local\Programs\Python\Python36\lib\site-packages\requests\sessions.py", line 462, in prepare_request
    hooks=merge_hooks(request.hooks, self.hooks),
  File "C:\Users\Islam\AppData\Local\Programs\Python\Python36\lib\site-packages\requests\models.py", line 313, in prepare
    self.prepare_url(url, params)
  File "C:\Users\Islam\AppData\Local\Programs\Python\Python36\lib\site-packages\requests\models.py", line 390, in prepare_url
    raise InvalidURL("Invalid URL %r: No host supplied" % url)
requests.exceptions.InvalidURL: Invalid URL 'https:https://i.annihil.us/u/prod/marvel/i/mg/6/b0/5b6c5e4154f75/portrait_uncanny.jpg': No host supplied

每当我在任何网站上尝试时，它都会给我。

Answer 1

它看起来像elem[0].get('src')评估为https://i.annihil.us/u/prod/marvel/i/mg/6/b0/5b6c5e4154f75/portrait_uncanny.jpg 。
因此，在comicurl='https:'+elem[0].get('src')您在一个格式正确的网址前添加了http: ，使其无效

Answer 2

不能对此提出Invalid URL 'https:https://i.annihil.us/u/prod ： Invalid URL 'https:https://i.annihil.us/u/prod该URL确实是无效的，可能您应该在以下语句中删除https ：

comicurl='https:'+elem[0].get('src')