CODEWITHSUNDEEP
pythonimage
J.Xi
J.Xi

Reputation: 1

Download a picture from certain URL with Python

I learned how to download a picture from a certain URL with python as:

import urllib
imgurl="http://www.digimouth.com/news/media/2011/09/google-logo.jpg"
resource = urllib.urlopen(imgurl)
output = open("test.jpg","wb")
output.write(resource.read())
output.close()

and it worked well, but when i changed the URL to 

  imgurl="http://farm1.static.flickr.com/96/242125326_607a826afe_o.jpg"

it did not work, and gave the information

File "face_down.py", line 3, in <module>
resource = urllib2.urlopen(imgurl)
File "D:\Python27\another\Lib\urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "D:\Python27\another\Lib\urllib2.py", line 431, in open
response = self._open(req, data)
File "D:\Python27\another\Lib\urllib2.py", line 449, in _open
'_open', req)
File "D:\Python27\another\Lib\urllib2.py", line 409, in _call_chain
result = func(*args)
File "D:\Python27\another\Lib\urllib2.py", line 1227, in http_open
return self.do_open(httplib.HTTPConnection, req)
File "D:\Python27\another\Lib\urllib2.py", line 1197, in do_open
raise URLError(err)
urllib2.URLError: <urlopen error [Errno 10060] >

and I tried to open the latter image URL, and it could be shown as the former, I have no idea to solve it~~ help~~~~

Upvotes: 0

Views: 425

Answers (2)

nishparadox
nishparadox

Reputation: 770

You can try using requests module. The response will be some bytes. So, you can iterate over those byte chunks and write to the file.

import requests

url = "http://farm1.static.flickr.com/96/242125326_607a826afe_o.jpg"
r = requests.get(url)
path = "filename.jpg"
with open(path, 'wb') as f:
    for chunk in r:
        f.write(chunk)

Upvotes: 1

Hello Name
Hello Name

Reputation: 37

I looked up both of the addresses and the second one does not lead anywhere. That is probably the problem. 

import urllib
imgurl="webpage url"
openimg = urllib.urlopen(imgurl) #opens image (prepares it)
img = open("test.jpg","wb") #opens the img to read it
img.write(openimg.read()) #prints it to the console
img.close() #closes img

Try the link again in your webpage and if it turns up with "webpage not available" that is probably the problem.

Upvotes: 0

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