| help | unix | grep (GNU grep) 2.5.1 | advanced regex syntax

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# 1  
Old 10-13-2009
| help | unix | grep (GNU grep) 2.5.1 | advanced regex syntax


I'm working on unix with grep (GNU grep) 2.5.1. I'm going through some of the newer regex syntax using Regular Expression Reference - Advanced Syntax a guide.

ls -aLl /bin | grep "\(x\)"

Which works, just highlights 'x' where ever, when ever.

I'm trying to to get (?:) to work but I just get no return.

 ls -aLl /bin | grep "\(\?\:x\)"

 ls -aLl /bin | grep "\(\?:x\)"

 ls -aLl /bin | grep "\(?\:x\)"

All of these return nothing. I'm not sure what I need to do to implement this. Like GNU grep supports the advance features like look ahead/behind, conditionals, backreference, etc?

The below code works which using grouping:
ls -aLl /bin | grep "^\([^x]*x\)\{1\}[^x]*$"

# 2  
Old 10-15-2009
I don't undestand what you problem is since it seems you found a way to solve it?
# 3  
Old 10-15-2009
Hello, I apologize if the message was unclear.

The problem is: I want to use the following advanced regular expression syntax "(?:)" which represents grouping without backreference.
Related information: From what I read GNU Grep supports the advanced syntax.
Support: I've used "()" which is grouping with backreference.

My goal isn't to solve any particular problem really. I'm new to the unix enviroment and right now I'm trying to learn as much about grep and regular expressions as possible. I understand that the advanced syntax with grep is likely going to be academic only since I would find piping into other programs simpler (sed/awk).

Summary: Can I use the syntax "(?:)" with GNU Grep 2.5.1?
# 4  
Old 10-15-2009
According to "GNU Extended Regular Expressions", the EREs are just like the usual, basic grep REs, only that special characters for grouping, repetition, ... don't need to be escaped. I've seen non-capturing groups, zero-width assertions, ... in Perl and the PCRE library so far (don't know about Python, Ruby, ... feel free to correct me).
# 5  
Old 10-15-2009
Ok, cool. You gave me some insight. For reference, the below line returns lines with exactly 3 matches if 'x' in the entire line.
3 can be replaced with any positive integer.
'x' can be replaced by any string.

grep "^\([^x]*x\)\{3\}[^x]*$" /usr/share/dict/words
grep -E "^([^x]*x){3}[^x]*$" /usr/share/dict/words

Addinging the option -E allowed to removed the "\" for the special characters and the code worked as expected.
There is also -P option for Perl extensions.

grep -E "^(?:[^x]*x){3}[^x]*$" /usr/share/dict/words
grep -P "^(?:[^x]*x){3}[^x]*$" /usr/share/dict/words

-E returned nothing when I added the ?: to ()
-P returned everything

I'm getting closer, I'll just need sometime to fool around with it. Thanks for the help.

Regular Expression Reference - Advanced Syntax
UNIX man pages : grep ()
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