urlretrieve get file name

[Gabriel Genellina]

At Friday 10/11/2006 16:58, Sven wrote:

>Yes the browser suggests a file name, but I did a little research using
>http://web-sniffer.net/. The Response Header contains roughly this:
>
>HTTP Status Code: HTTP/1.1 302 Found
>Location: http://page.com/filename.zip
>Content-Length: 0
>Connection: close
>Content-Type: text/html
>
>The status code 302 tells the browser where to find the file. The funny
>thing is that calling the info() function, on the file-like response
>object, in Python doesn't return the same header. I'm so stuck. :-)
>Thanks for your help.

Because urlopen is smart enough to detect the redirection and do a 
second request.
You can use the geturl() method to obtain the true URL used (that 
would be http://page.com/filename.zip) and then rename the file.
Or, you can install your own URLOpener (I think a FancyURLOpener with 
retries=0 would be OK) and process the Location header yourself. See 
the urllib documentation.


-- 
Gabriel Genellina
Softlab SRL 

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