On Apr 7, 1:44 pm, Tim Chase wrote:
> > f = urllib.urlopen("http://www.google.com";)
> > s = f.read()
>
> > It is working, but it's returning the source of the page. Is there anyway I
> > can get almost a screen capture of the page?
>
> This is the job of a browser -- to render the source HTML. A
In message , Support
Desk wrote:
> You could do something like below to get the rendered page.
>
> Import os
> site = 'website.com'
> X = os.popen('lynx --dump %s' % site).readlines()
I wonder how easy it would be to get the page image in SVG format? I believe
the Gecko HTML engine in Firefox
> Is there anyway I
> can get almost a screen capture of the page?
I'm not sure exactly what you mean by "screen capture". But the
webbrowser module in the standard lib might be of some help. You can
use it to drive a web browser from Python.
to load a page in your browser, you can do something
From: Ronn Ross [mailto:ronn.r...@gmail.com]
Sent: Tuesday, April 07, 2009 9:37 AM
To: Support Desk
Subject: Re: Scraping a web page
This works great, but is there a way to do this with firefox or something
similar so I can also print the images from the site?
On Tue, Apr 7, 2009 at 9:58
Ronn Ross
Cc: python-list@python.org
Subject: Re: Scraping a web page
> f = urllib.urlopen("http://www.google.com";)
> s = f.read()
>
> It is working, but it's returning the source of the page. Is there anyway
I
> can get almost a screen capture of the page?
This is the
f = urllib.urlopen("http://www.google.com";)
s = f.read()
It is working, but it's returning the source of the page. Is there anyway I
can get almost a screen capture of the page?
This is the job of a browser -- to render the source HTML. As
such, you'd want to look into any of the browser-aut
