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sed : regular expression (hungry mode ?) solved

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# 1  
Old 11-15-2010
sed : regular expression (hungry mode ?) solved
Hi
there is something I don't understand with the repeated element. With the following command I have a weird output:
Quote:
$ echo 3_aaaun+g | sed "s/.*\(aa*\).*/>\1</g"
>a<
Smilie
while I thought the output would be
Quote:
>aaa<
I really don't understand why 2 "a" are skipped. Might someone explain please?
# 2  
Old 11-15-2010
That is because of sed's greedy matching. The .* matches as much as possible so that \(aa*\) only has one a left (the rightmost a). This should remedy it:
Code:
sed "s/[^a]*\(aa*\).*/>\1</g"

or
Code:
sed "s/.*_\(aa*\).*/>\1</g"

This User Gave Thanks to Scrutinizer For This Post:
xib.be
# 3  
Old 11-15-2010
The first .* is eating some of the a's. If you don't want it to eat a's, then give it [icode][^a]* instead of .*

And if you always want it to start at the beginning and end at the end of the string, you should pin it in those places too.

Code:
$  echo 3_aaaun+g | sed "s/^[^a]*\(aa*\).*$/>\1</g"
>aaa<
$

This User Gave Thanks to Corona688 For This Post:
xib.be
# 4  
Old 11-15-2010
Thank you both of you Smilie so fast reply ^^ I like unix.com Smilie
Scrutinizer you soleved my trouble.

But just to know is there a way to remove the "greedy mode" ? ... in the man I found nothing :s
# 5  
Old 11-15-2010
There is no lazy matching in sed, but by using [^a] (any character but a) instead of . (any character) you can go a long way. Otherwise you would have to switch to perl:
Code:
perl -pe 's/.*?(a+).*/>\1</g'

Last edited by Scrutinizer; 11-15-2010 at 06:58 PM..
This User Gave Thanks to Scrutinizer For This Post:
xib.be
# 6  
Old 11-16-2010
thank you again Scrutinizer Smilie
perl will get me sooner or later x)

Edit: Hmm I can't "close" my topic ? or put it in a solved mode? Smilie
Last edited by jim mcnamara; 11-16-2010 at 08:22 AM.. Reason: Not closed, marked 'solved'
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