urlretrieve get file name

[Sven]

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 point me in the right direction are greatly appreciated.

Sven

 

[Gabriel Genellina]

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 point me in the right direction are greatly appreciated.

The file name *may* come in the Content-Disposition header (ex:
Content-Disposition: attachment; filename="budget.xls")
Use urlopen to obtain a file-like object; its info() method gives you
those headers.


--
Gabriel Genellina
Softlab SRL

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Correo Yahoo!
Espacio para todos tus mensajes, antivirus y antispam ¡gratis!
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[Sven]

Hello Gabriel,

Thanks for your help, but I'm a guy with no luck. I can't get the
file name from response header...

 

[Gabriel Genellina]

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 filen ame, the server is not providing one
(there is no obligation to do so).


--
Gabriel Genellina
Softlab SRL

__________________________________________________
Correo Yahoo!
Espacio para todos tus mensajes, antivirus y antispam ¡gratis!
¡Abrí tu cuenta ya! - http://correo.yahoo.com.ar

 

[Sven]

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.

 

[Gabriel Genellina]

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

__________________________________________________
Correo Yahoo!
Espacio para todos tus mensajes, antivirus y antispam ¡gratis!
¡Abrí tu cuenta ya! - http://correo.yahoo.com.ar

 

[Sven]

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.