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Groups > comp.lang.python > #9795 > unrolled thread
| Started by | srikanth <srikanth007m@gmail.com> |
|---|---|
| First post | 2011-07-18 05:05 -0700 |
| Last post | 2011-07-18 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
| From | srikanth <srikanth007m@gmail.com> |
|---|---|
| Date | 2011-07-18 05:05 -0700 |
| Subject | open 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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| From | Chris Angelico <rosuav@gmail.com> |
|---|---|
| Date | 2011-07-18 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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| From | srikanth <srikanth007m@gmail.com> |
|---|---|
| Date | 2011-07-18 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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