当我尝试这一行时:
import urllib.request
urllib.request.urlretrieve("https://i.redd.it/53tfh959wnv41.jpg", "photo.jpg")
我得到以下错误:
Traceback (most recent call last):
File "scraper.py", line 26, in <module>
urllib.request.urlretrieve("https://i.redd.it/53tfh959wnv41.jpg", "photo.jpg")
File "/usr/lib/python3.6/urllib/request.py", line 248, in urlretrieve
with contextlib.closing(urlopen(url, data)) as fp:
File "/usr/lib/python3.6/urllib/request.py", line 223, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python3.6/urllib/request.py", line 532, in open
response = meth(req, response)
File "/usr/lib/python3.6/urllib/request.py", line 642, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.6/urllib/request.py", line 570, in error
return self._call_chain(*args)
File "/usr/lib/python3.6/urllib/request.py", line 504, in _call_chain
result = func(*args)
File "/usr/lib/python3.6/urllib/request.py", line 650, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found
但是链接在我的浏览器中运行良好?为什么它可以在浏览器中工作,但不能用于请求?它与来自同一站点的其他图片一起工作
尝试更改用户代理。您只需添加一个kwarg:
请求返回
如果您检查开发人员控制台,它是404:
因此,您看到的是imgur的自定义404“页面”(这是一个图像)
编辑:
因此
urlretrieve
在404状态代码上失败。如果要使用请求的内容(即使状态代码为404),可以执行以下操作:相关问题 更多 >
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