> You can use the geturl() method to obtain the true URL used (that
> would behttp://page.com/filename.zip) and then rename the file.
Thanks mate, this was exactly what I needed. A realy clean and simple
solution to my problem. :-)
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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
Con
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 t
At Thursday 9/11/2006 20:52, Sven wrote:
Thanks for your help, but I'm a guy with no luck. :-) I can't get the
file name from response header...
Try using a browser and "Save as..."; if it suggests a file name, it
*must* be in the headers - so look again carefully.
If it does not suggests a f
Hello Gabriel,
Thanks for your help, but I'm a guy with no luck. :-) I can't get the
file name from response header...
On Nov 10, 12:39 am, Gabriel Genellina <[EMAIL PROTECTED]> wrote:
> At Thursday 9/11/2006 19:11, Sven wrote:
>
> >I'm wrestling with the urlretrieve function in the urllib module
At Thursday 9/11/2006 19:11, Sven wrote:
I'm wrestling with the urlretrieve function in the urllib module. I
want to download a file from a web server and save it locally with the
same name. The problem is the URL - it's on the form
http://www.page.com/?download=12345. It doesn't reveal the file
Hi guys and gals,
I'm wrestling with the urlretrieve function in the urllib module. I
want to download a file from a web server and save it locally with the
same name. The problem is the URL - it's on the form
http://www.page.com/?download=12345. It doesn't reveal the file name.
Some hints to poin
