Re: [SLUG] Perl Regular expression help
Sorry to bring up an old thread but I just had to comment on this... $quoted_author = Jamie Wilkinson ; Try: /pg=[^]*/ match zero or more of the character class that is not an ampersand. Except there is nothing stopping the variables being reordered, no? So you may need to match a leading ? instead of . You could get crazy and try to do this in a single regex but two stage is clearer. e.g. sed -e 's/pg=[^]*//g' -e 's/?pg=[^]*/?/' cheers Marty -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
Call me crazy! s/(|?)pg=[^]*/\1/ (correct escaping of and ? left as an exercise for someone actually using this :) On 27 July 2010 08:03, Martin Barry ma...@supine.com wrote: Sorry to bring up an old thread but I just had to comment on this... $quoted_author = Jamie Wilkinson ; Try: /pg=[^]*/ match zero or more of the character class that is not an ampersand. Except there is nothing stopping the variables being reordered, no? So you may need to match a leading ? instead of . You could get crazy and try to do this in a single regex but two stage is clearer. e.g. sed -e 's/pg=[^]*//g' -e 's/?pg=[^]*/?/' cheers Marty -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
On 28/07/2010, at 1:03, Martin Barry ma...@supine.com wrote: You could get crazy and try to do this in a single regex but two stage is clearer. e.g. sed -e 's/pg=[^]*//g' -e 's/?pg=[^]*/?/' Now you have 2 problems. -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Fwd: Re: [SLUG] Perl Regular expression help
Perhaps this is the time for me to ask: does anyone know a way for grep or awk to extract from a text file any sequence of up to, say, six words that begins and ends with an initially-capitalised word -- whether or not it is part of a larger matching sequence? So if the input text was: Sally Lee Jones worked for the United Nations Support Team the output would be Sally Lee Lee Jones Sally Lee Jones Jones worked for the United Jones worked for the United Nations United Nations Nations Support Support Team United Nations Support Team I don't particularly care if it takes one pass or several, and I can clean up duplicates afterwards. This is not a serious problem for me -- it falls into the 'would be nice to have' category -- but I've been puzzling over it off and on for a while, and the mention of the word 'greedy' reminded me of it. Jon. On 14/07/10 18:06, Nick Andrew wrote: (aaa...) Where the stuff inside () is what's being matched. The matched part stops at the first or the end of the string. It's greedy so it matches as long a string as possible. Nick. -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
On Wed, Jul 14, 2010 at 01:27:13PM +1000, Peter Rundle wrote: I don't really understand how the [^] followed by the * works but it does. any character which is not an ampersand repeated zero or more times. So it matches () () (a) (a) (aaa...) (aaa...) Where the stuff inside () is what's being matched. The matched part stops at the first or the end of the string. It's greedy so it matches as long a string as possible. Nick. -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
On 14/07/2010, at 13:27, Peter Rundle pe...@aerodonetix.com.au wrote: P.S I didn't understand Lindsay's question about doing the replace. I'm replacing the arg with nothing, I.E I just want to remove the pg= argument from the string. Didn't know what you were replacing your match with, was just curious about how other people would solve this problem. Lindsay -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
Thanks Nick, What I didn't know was that ^ inside brackets [] means not. I was still reading ^ as beginning of string. That will be very very useful in future. Thanks Pete Nick Andrew wrote: On Wed, Jul 14, 2010 at 01:27:13PM +1000, Peter Rundle wrote: I don't really understand how the [^] followed by the * works but it does. any character which is not an ampersand repeated zero or more times. So it matches () () (a) (a) (aaa...) (aaa...) Where the stuff inside () is what's being matched. The matched part stops at the first or the end of the string. It's greedy so it matches as long a string as possible. Nick. -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
Try: /pg=[^]*/ match zero or more of the character class that is not an ampersand. On 13 July 2010 17:21, Peter Rundle pe...@aerodonetix.com.au wrote: Hi Sluggers, I'm sure some of you genii have a real quick solution to this. I'm trying to find and replace and argument in a url. The url is of the form pg=somethingarg=somethingelse I want to take out the pg=something but the arg= may or may not be there. How do I say match the pg=something up to but not including the next (which may or may not be there). /pg=.*/ But also I think is a special char (no?) that means put the matched bit back, though is that only on the replace side? (my question relates strictly to the matching side). TIA's Pete -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
Now you've got the search, I'm curious how you are going to do the replace. Is the Perlism to just use the substitute operator, or split on the pattern, iterate through the array, and join again? Lindsay On 14 July 2010 10:30, Jamie Wilkinson j...@spacepants.org wrote: Try: /pg=[^]*/ match zero or more of the character class that is not an ampersand. On 13 July 2010 17:21, Peter Rundle pe...@aerodonetix.com.au wrote: Hi Sluggers, I'm sure some of you genii have a real quick solution to this. I'm trying to find and replace and argument in a url. The url is of the form pg=somethingarg=somethingelse I want to take out the pg=something but the arg= may or may not be there. How do I say match the pg=something up to but not including the next (which may or may not be there). /pg=.*/ But also I think is a special char (no?) that means put the matched bit back, though is that only on the replace side? (my question relates strictly to the matching side). TIA's Pete -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- w: http://holmwood.id.au/~lindsay/ t: @auxesis -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
I'd use a global search and replace command, if it were me, and I was using sed: sed -ie 's/pg=[^]//g' lindsay.html On 13 July 2010 18:13, Lindsay Holmwood lind...@holmwood.id.au wrote: Now you've got the search, I'm curious how you are going to do the replace. Is the Perlism to just use the substitute operator, or split on the pattern, iterate through the array, and join again? Lindsay On 14 July 2010 10:30, Jamie Wilkinson j...@spacepants.org wrote: Try: /pg=[^]*/ match zero or more of the character class that is not an ampersand. On 13 July 2010 17:21, Peter Rundle pe...@aerodonetix.com.au wrote: Hi Sluggers, I'm sure some of you genii have a real quick solution to this. I'm trying to find and replace and argument in a url. The url is of the form pg=somethingarg=somethingelse I want to take out the pg=something but the arg= may or may not be there. How do I say match the pg=something up to but not including the next (which may or may not be there). /pg=.*/ But also I think is a special char (no?) that means put the matched bit back, though is that only on the replace side? (my question relates strictly to the matching side). TIA's Pete -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- w: http://holmwood.id.au/~lindsay/ t: @auxesis -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
how about using a slightly different approach with split @input = split /\/; $input[0] should now be pg=something, $input[1] will be the args=somthingelse so you can trivially match, modify and print this to your output, whether or not it has extra arguments. On Wed, Jul 14, 2010 at 11:24 AM, Jamie Wilkinson j...@spacepants.orgwrote: I'd use a global search and replace command, if it were me, and I was using sed: sed -ie 's/pg=[^]//g' lindsay.html On 13 July 2010 18:13, Lindsay Holmwood lind...@holmwood.id.au wrote: Now you've got the search, I'm curious how you are going to do the replace. Is the Perlism to just use the substitute operator, or split on the pattern, iterate through the array, and join again? Lindsay On 14 July 2010 10:30, Jamie Wilkinson j...@spacepants.org wrote: Try: /pg=[^]*/ match zero or more of the character class that is not an ampersand. On 13 July 2010 17:21, Peter Rundle pe...@aerodonetix.com.au wrote: Hi Sluggers, I'm sure some of you genii have a real quick solution to this. I'm trying to find and replace and argument in a url. The url is of the form pg=somethingarg=somethingelse I want to take out the pg=something but the arg= may or may not be there. How do I say match the pg=something up to but not including the next (which may or may not be there). /pg=.*/ But also I think is a special char (no?) that means put the matched bit back, though is that only on the replace side? (my question relates strictly to the matching side). TIA's Pete -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- w: http://holmwood.id.au/~lindsay/ t: @auxesis -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
RE: [SLUG] Perl Regular expression help
/pg=.*/ But also I think is a special char (no?) that means put the matched bit back, though is that only on the replace side? (my question relates strictly to the matching side). Yes the ampersand is special, it represents the complete matched string on the replace. s/pg=.*/\/ As pointed out the solution is not optimal, if there is more than two parameters it will consume them all. It will also NOT remove a trailing parameter because the second is not there. Ken -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
I don't see the problem with my approach; the match will terminate when it sees the second ampersand, without consuming it. On 13 July 2010 19:01, Ken Foskey kfos...@tpg.com.au wrote: /pg=.*/ But also I think is a special char (no?) that means put the matched bit back, though is that only on the replace side? (my question relates strictly to the matching side). Yes the ampersand is special, it represents the complete matched string on the replace. s/pg=.*/\/ As pointed out the solution is not optimal, if there is more than two parameters it will consume them all. It will also NOT remove a trailing parameter because the second is not there. Ken -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
Re: [SLUG] Perl Regular expression help
Thanks Jamie (and others), That works a treat. I would have tried /pg=*[^]/ which of course would have matched all ampersands up until the last taking out more than one argument. I don't really understand how the [^] followed by the * works but it does. Thanks Pete P.S I didn't understand Lindsay's question about doing the replace. I'm replacing the arg with nothing, I.E I just want to remove the pg= argument from the string. Jamie Wilkinson wrote: Try: /pg=[^]*/ match zero or more of the character class that is not an ampersand. -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
