Negate matching group in perl

[wojtek75]

Hi,

there is a text like that:
start in
start out
start off
start in
start away
start with
...

I like to match lines:
start (in|out)
but don't like to match lines with all other prepositions:
start (off|away|back|with|...)

How can I implement it in perl? The '^' doesn't seem to negate the () grouping and
the following doesn't work:
m/start\s^(in|out)/

I would appreciate your help with it.

[James R. Ferguson · Accepted solution]

Hi:

To match 'start (in|out)' you could simply do:

# perl -ne 'm{start\s+(?:in|out)} and print' file

...the (?:) gives grouping-only without capture.

You could also do:

# perl -ne 'm{start\s+(?=in|out)} and print' file

...where the (?=) denotes a positive look-ahead.

And then:

# perl -ne 'm{start\s+(?!in|out)} and print' file

...which is a negative look-ahead which yields what you seek.

Regards!

...JRF...

[wojtek75]

Thanks. Is such feature available in egrep command?

[H.Merijn Brand (procura]

Only if egrep is built with PCRE support

(mine is)

in which case grep knows the option -P and expressions are interpreted perl-style

Enjoy, Have FUN! H.Merijn

[wojtek75]

Thanks,

and yet another perl challange. If my file includes "start " phrase anywhere inside I want to return false. The only acceptable exception is exact "start with" phrase. Bear in mind that ie. "start without" should return false either as "without" matches here.

Thanks in advance for sharing your expertise.

[James R. Ferguson]

Hi (again):

In keeping with yesterday's thread:

# perl -nl -0377 -e 'print m{start\s+with\b}?1:0' file

Regards!

...JRF...

[wojtek75]

Thanks,

I think I was not too specific. This is what I need in meta. is a template for any word.

if scirpts includes "start " phrase then {
if anyword is "with" then {
return 1
} else {
return 0
}
} else {
return 1
}

The goal is to find files with "start " which I find risky. The only acceptable exception is "start with" that I find safe.
Your solution returns false even when no "start " clause is there.

[James R. Ferguson]

Hi (again):

Per your last comments, you might do:

# perl -ne '$n++ if m{start\s+(?!with\b)};exit 1 if (m{start\s+with\b} and $n)' file;echo $?

Regards!

...JRF...

[wojtek75]

Thanks,

now I isolated my problem to the following:

if [[ $(perl -nl -0377 -e 'print m{start[\s\n]+(?!with\b)}i;' a.txt) -gt 0 ]]; then
print "Start non-with"
else
print "Start with"
fi

The input files are as follows (_ is for space):
1. =======
start with
2. =======
start
with
3. =======
start_
with
4. =======
start

with
======

1st and 2nd input returns "Start with" message. Why 3rd and 4th don't?

[James R. Ferguson]

Hi:

Let's simplify things. You're looking for the sequence "start with" so:

if [[ -z "$(perl -nl -0377 -e 'print m{start\s+with\b}i' ${FILE})" ]]; then
print "Start non-with"
else
print "Start with"
fi

...

Nothing will be returned if no match is found, hence, we will compare (in the shell) for an empty string.

Regards!

...JRF...

[wojtek75]

Frankly speaking I am looking for "start " while "start with" is an the only acceptable exception. Any number of spaces and newlines may be between "start" and "".

BTW, are you able to explain why my previous regexp doesn't work with 3rd and 4th input?

[James R. Ferguson]

HI (again):

OK, I think I understand where we left off. Try this:

#!/usr/bin/sh
typeset FILE=$1
if [[ $(perl -0777 -nle 'print m{start\s+(?!\s*with\b)}i?1:0' ${FILE}) -gt 0 ]]; then
print "Start non-with"
else
print "Start with"
fi

...I bungled and mislead you when I erroneously used '-0377'. I meant '-0777' and was thinking of something else. This lead to the match errors.

Regards!

...JRF...

[James R. Ferguson]

Hi (again):

I'm not thinking clearly today :-(

Use:

if [[ $(perl -0777 -nle 'print m{start(?!\s+with\b)}i?1:0' ${FILE}) -gt 0 ]]; then
...

If this or the immediately prior post of mine helps, assign 0-points to the first one as this is what I should have offered!

Regards!

...JRF...