问题
I'm writing a small Python script to grab images via google images. I've managed to get things up to the point where I have the urls of the images I want in a handy list. Now, I just need to grab them...
for each image url i do this:
print("Retrieving:{0}".format(sFinalImageURL))
sExt = sFinalImageURL.split('.')[-1]
#u = urllib.request.urlopen(sFinalImageURL)
try:
u = urllib.request.urlopen(sFinalImageURL)
except:
print("error: cannot retrieve image")
continue
raw_data = u.read()
print("read {0} bytes".format(len(raw_data)))
u.close()
global sImagesFolder
try:
f = open("{0}/{1}_{2}.{3}".format(sImagesFolder,sImage,i,sExt),'wb')
f.write(raw_data)
f.close()
except:
print("couldn't write to {0}/{1}_{2}.{3}".format(sImagesFolder,sImage,i,sExt))
print()
Here are the problems I'm coming up against:
trying to open some off the URLs gives me 403 even though I can open the URLs straight in my browser. So there's something in the HTTP request header that the image server doesn't like... any ideas?
Here's some of the output:
Retrieving:http://upload.wikimedia.org/wikipedia/commons/thumb/4/43/Timba%2B1.jpg/220px-Timba%2B1.jpg
error: cannot retrieve image
Retrieving:http://upload.wikimedia.org/wikipedia/commons/thumb/2/26/YellowLabradorLooking_new.jpg/260px-YellowLabradorLooking_new.jpg
error: cannot retrieve image
Retrieving:http://1.bp.blogspot.com/-7SsJ1n3RdoA/Tf07NOgD5nI/AAAAAAAAABo/tl8qLLIU01Y/s1600/english-shepherd-dog-0003.jpg
read 11123 bytes
Retrieving:http://completedogfood.net/wp-content/uploads/2010/07/complete-dog-food.bmp
read 419630 bytes
回答1:
It seems like Wikipedia only allows access to real browsers.
The problem can be solved by specifying a User-Agent
string of a real browser, because Python's urllib
sends something like Python-urllib/3.2
by default.
Here's an example that works (with User-Agent
string of the browser that I use):
url = 'http://upload.wikimedia.org/wikipedia/commons/thumb/4/43/Timba%2B1.jpg/220px-Timba%2B1.jpg'
user_agent = 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/535.19 (KHTML, like Gecko) Ubuntu/12.04 Chromium/18.0.1025.168 Chrome/18.0.1025.168 Safari/535.19'
u = urllib.request.urlopen(urllib.request.Request(url, headers={'User-Agent': user_agent}))
来源:https://stackoverflow.com/questions/10945542/python3-urllib-image-retreval