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Groups > comp.lang.python > #9795 > unrolled thread

open urls in browser

Started bysrikanth <srikanth007m@gmail.com>
First post2011-07-18 05:05 -0700
Last post2011-07-18 06:36 -0700
Articles 3 — 2 participants

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  open urls in browser srikanth <srikanth007m@gmail.com> - 2011-07-18 05:05 -0700
    Re: open urls in browser Chris Angelico <rosuav@gmail.com> - 2011-07-18 23:21 +1000
      Re: open urls in browser srikanth <srikanth007m@gmail.com> - 2011-07-18 06:36 -0700

#9795 — open urls in browser

Fromsrikanth <srikanth007m@gmail.com>
Date2011-07-18 05:05 -0700
Subjectopen urls in browser
Message-ID<39758958-fc2f-4114-a420-d41ce978e277@r28g2000prb.googlegroups.com>
Hi All,
I am new to python. Before posting i have done some google regarding
my question. But i didn't get exact information. So thought of
posting
it here. I want to open a list of urls in browser that too in same
window with out exiting. it should load one by one on same window and
also it should log the details simultaneously to another text file.
Ex: http://www.google.com - Pass/Fail. If you guys already having
this
code please share it with me. Thanks a lot in advance.

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#9802

FromChris Angelico <rosuav@gmail.com>
Date2011-07-18 23:21 +1000
Message-ID<mailman.1217.1310995298.1164.python-list@python.org>
In reply to#9795
On Mon, Jul 18, 2011 at 10:05 PM, srikanth <srikanth007m@gmail.com> wrote:
> Ex: http://www.google.com - Pass/Fail.

What do you mean by "Pass" or "Fail"? If you send a URL to a web
browser, all you'll find out is whether or not the browser accepted it
- it won't tell you if the page is valid. If you want that, you don't
need a web browser at all - what you want is a simple URL fetcher,
such as urllib/urllib2.

http://docs.python.org/library/urllib.html
http://docs.python.org/release/3.1.3/library/urllib.request.html

ChrisA

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#9803

Fromsrikanth <srikanth007m@gmail.com>
Date2011-07-18 06:36 -0700
Message-ID<e05a09d4-db0d-4a25-90cd-a5e34f6b69ef@v11g2000prn.googlegroups.com>
In reply to#9802
On Jul 18, 6:21 pm, Chris Angelico <ros...@gmail.com> wrote:
> On Mon, Jul 18, 2011 at 10:05 PM, srikanth <srikanth0...@gmail.com> wrote:
> > Ex:http://www.google.com- Pass/Fail.
>
> What do you mean by "Pass" or "Fail"? If you send a URL to a web
> browser, all you'll find out is whether or not the browser accepted it
> - it won't tell you if the page is valid. If you want that, you don't
> need a web browser at all - what you want is a simple URL fetcher,
> such as urllib/urllib2.
>
> http://docs.python.org/library/urllib.htmlhttp://docs.python.org/release/3.1.3/library/urllib.request.html
>
> ChrisA

Sorry i don't need that one. by mistake i have written it.

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