Downloading a picture via urllib and python.

Question

So I'm trying to make a Python script that downloads webcomics and puts them in a folder on my desktop. I've found a few similar programs on here that do something similar, but nothing quite like what I need. The one that I found most similar is right here (http://bytes.com/topic/python/answers/850927-problem-using-urllib-download-images). I tried using this code:

>>> import urllib
>>> image = urllib.URLopener()
>>> image.retrieve("http://www.gunnerkrigg.com//comics/00000001.jpg","00000001.jpg")
('00000001.jpg', <httplib.HTTPMessage instance at 0x1457a80>)

I then searched my computer for a file "00000001.jpg", but all I found was the cached picture of it. I'm not even sure it saved the file to my computer. Once I understand how to get the file downloaded, I think I know how to handle the rest. Essentially just use a for loop and split the string at the '00000000'.'jpg' and increment the '00000000' up to the largest number, which I would have to somehow determine. Any reccomendations on the best way to do this or how to download the file correctly?

Thanks!

EDIT 6/15/10

Here is the completed script, it saves the files to any directory you choose. For some odd reason, the files weren't downloading and they just did. Any suggestions on how to clean it up would be much appreciated. I'm currently working out how to find out many comics exist on the site so I can get just the latest one, rather than having the program quit after a certain number of exceptions are raised.

import urllib
import os

comicCounter=len(os.listdir('/file'))+1  # reads the number of files in the folder to start downloading at the next comic
errorCount=0

def download_comic(url,comicName):
    """
    download a comic in the form of

    url = http://www.example.com
    comicName = '00000000.jpg'
    """
    image=urllib.URLopener()
    image.retrieve(url,comicName)  # download comicName at URL

while comicCounter <= 1000:  # not the most elegant solution
    os.chdir('/file')  # set where files download to
        try:
        if comicCounter < 10:  # needed to break into 10^n segments because comic names are a set of zeros followed by a number
            comicNumber=str('0000000'+str(comicCounter))  # string containing the eight digit comic number
            comicName=str(comicNumber+".jpg")  # string containing the file name
            url=str("http://www.gunnerkrigg.com//comics/"+comicName)  # creates the URL for the comic
            comicCounter+=1  # increments the comic counter to go to the next comic, must be before the download in case the download raises an exception
            download_comic(url,comicName)  # uses the function defined above to download the comic
            print url
        if 10 <= comicCounter < 100:
            comicNumber=str('000000'+str(comicCounter))
            comicName=str(comicNumber+".jpg")
            url=str("http://www.gunnerkrigg.com//comics/"+comicName)
            comicCounter+=1
            download_comic(url,comicName)
            print url
        if 100 <= comicCounter < 1000:
            comicNumber=str('00000'+str(comicCounter))
            comicName=str(comicNumber+".jpg")
            url=str("http://www.gunnerkrigg.com//comics/"+comicName)
            comicCounter+=1
            download_comic(url,comicName)
            print url
        else:  # quit the program if any number outside this range shows up
            quit
    except IOError:  # urllib raises an IOError for a 404 error, when the comic doesn't exist
        errorCount+=1  # add one to the error count
        if errorCount>3:  # if more than three errors occur during downloading, quit the program
            break
        else:
            print str("comic"+ ' ' + str(comicCounter) + ' ' + "does not exist")  # otherwise say that the certain comic number doesn't exist
print "all comics are up to date"  # prints if all comics are downloaded

Answer 1

It's easiest to just use .read() to read the partial or entire response, then write it into a file you've opened in a known good location.

Answer 2

Aside from suggesting you read the docs for retrieve() carefully (http://docs.python.org/library/urllib.html#urllib.URLopener.retrieve), I would suggest actually calling read() on the content of the response, and then saving it into a file of your choosing rather than leaving it in the temporary file that retrieve creates.

Answer 3

import urllib
f = open('00000001.jpg','wb')
f.write(urllib.urlopen('http://www.gunnerkrigg.com//comics/00000001.jpg').read())
f.close()

Answer 4

Using urllib.urlretrieve:

import urllib
urllib.urlretrieve("http://www.gunnerkrigg.com//comics/00000001.jpg", "00000001.jpg")