I'm using urlretrieve from the urllib module.
I cannot seem to find how to add 开发者_如何学JAVAa User-Agent description to my requests.
Is it possible with urlretrieve? or do I need to use another method?
First, set version:
urllib.URLopener.version = 'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.153 Safari/537.36 SE 2.X MetaSr 1.0'
Then:
filename, headers = urllib.urlretrieve(url)
I know this issue had been there for 7 years. And I reached this issue by trying to figure out how to change the User-Agent
while using urlretrieve
function.
To anyone who reached this issue by no luck, here is how I did:
# proxy = ProxyHandler({'http': 'http://192.168.1.31:8888'})
proxy = ProxyHandler({})
opener = build_opener(proxy)
opener.addheaders = [('User-Agent','Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_4) AppleWebKit/603.1.30 (KHTML, like Gecko) Version/10.1 Safari/603.1.30')]
install_opener(opener)
result = urlretrieve(url=file_url, filename=file_name)
The reason I added proxy is to monitor the traffic in Charles, and here is the traffic I got:
You can use URLopener or FancyURLopener classes. The 'version' argument specifies the user agent of the opener object.
opener = FancyURLopener({})
opener.version = 'Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/47.0.2526.69 Safari/537.36'
opener.retrieve('http://example.com', 'index.html')
I don't think it's possible with urlretrieve - at least not easily. I would propose to create an urllib2.Request object and pass the required headers to it. See
http://docs.python.org/library/urllib2.html#urllib2.urlopen
for examples.
Something like this not using urllib tho, worked for me for an scraper
import requests
imageURL='http://image.jpg'
headers={'user-agent': 'Mozilla/5.0'}
r=requests.get(imageURL, headers=headers)
with open('image.jpg', 'wb') as f:
f.write(r.content)
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