topic Re: grep -E (with regular expression) \b in Operating System - HP-UX

grep -E (with regular expression) \b

Reggie Chang — Fri, 12 Mar 2004 16:30:34 GMT

Hi there,

I'm trying to use grep -E (regular expression)
with \b (word boundary) in csh without success.
(grep -E $path '\b\/usr....\b')

Is there any way to use \b, \s etc with grep?

Thanks.

Reggie

Re: grep -E (with regular expression) \b

curt larson_1 — Fri, 12 Mar 2004 17:02:13 GMT

what version of grep are you using, what /usr/bin/grep?

at 11.11 grep has the -w option that matches only words

Re: grep -E (with regular expression) \b

Reggie Chang — Fri, 12 Mar 2004 17:10:04 GMT

Curt,

I'm using 10.20. I don't know what
version if grep it is, but it is located
at /usr/bin.

Thanks.

Reggie

Re: grep -E (with regular expression) \b

Bill Hassell — Fri, 12 Mar 2004 17:15:39 GMT

grep at 10.20 does not have the -w (word boundary) option. You'll have to upgrade to 11.11 with the latest patches to have that feature. \b is not a part of regular expressions so grep can't parse it.

Re: grep -E (with regular expression) \b

Reggie Chang — Fri, 12 Mar 2004 17:22:40 GMT

Bill,

Thank you for the info.

I did not see -w in grep man page.
I'll find some other way since I'm
stuck with 10.20 at this moment.

Thanks.

Reggie

Re: grep -E (with regular expression) \b

curt larson_1 — Fri, 12 Mar 2004 17:23:26 GMT

you could also use the blank or space character classes

grep -E '[:space:]...[:space:]'
or
grep -E '[:blank:]...[:blank:]'

not even the -w switch is going to work for you in this case because both : and / are non word characters

your going to have to use perl or write your own

function notInPath {
typeset field
print $2 |
awk -v field=$1 -F: '{
if (NR > 1) exit;
for ( i=1; i<=NF; i++ ) {
if ( field == $i ) exit 1;
}
exit 0;}'
return
}

notInPath /usr/bin $PATH
if [[ $? = 0 ]] ;then
print "not in path"
else
print "it is in the path"
fi

Re: grep -E (with regular expression) \b

Reggie Chang — Fri, 12 Mar 2004 21:20:12 GMT

Curt,

Thanks a lot for the info.

Please forgive my dumb questions.
Is function notInPath in Perl?
Does shell script (C-shell or others)
support function?

Best regards,

Reggie

Re: grep -E (with regular expression) \b

curt larson_1 — Sat, 13 Mar 2004 00:55:18 GMT

the syntax for the function i provide is written for the posix shell or the korn shell. A portion of it could be call an awk script.

still using posix shell syntax, but using perl instead of awk

print "$PATH" |
/opt/perl/bin/perl -F: -ane '
foreach $str (@F) {
if ( $str =~ m|^/usr/bin$| ) {
exit 1;
}};exit 0;'

if [[ $? = 0 ]] ;then
print "not in path"
else
print "it is in the path"
fi

the above example doesn't use a function

Re: grep -E (with regular expression) \b

curt larson_1 — Sat, 13 Mar 2004 01:14:10 GMT

another easy way would be

print $PATH |
tr ":" "\012" |
grep -x /usr/bin >/dev/null 2>&1

if [[ $? = 0 ]] ;then
print "/usr/bin is in \$PATH"
else
print "/usr/bin is not in \$PATH"
fi

of if you like to switch the logic

num=$(print $PATH |
tr ":" "\012" |
grep -cx /usr/bin >/dev/null 2>&1)

if [[ $num = 0 ]] ;then
print "/usr/bin is not in \$PATH"
else
print "/usr/bin is in \$PATH"
fi

Re: grep -E (with regular expression) \b

Mark Grant — Sat, 13 Mar 2004 05:36:07 GMT

I might be being a bit dim here but doesn't this do what you want

echo $path|perl -ne "if(/\b\/usr...\b/){print}"

Re: grep -E (with regular expression) \b

john korterman — Sat, 13 Mar 2004 06:18:19 GMT

Hi,
if you have to use a regular expression you probably have to adapt it to your needs. The following is just a suggestion:
# grep -E "(^|[^a-zA-Z0-9])ordet([^a-zA-Z0-9]|$)" infile

which will try to look for the literal string
ordet
in infile. It does not take boundaries like e.g. special characters or tabs into account, but has already developed into something that gives unix a reputation as not easily understandable.....

regards,
John K.

Re: grep -E (with regular expression) \b

curt larson_1 — Sat, 13 Mar 2004 10:08:51 GMT

Mr. Grant,

in this case the data being compared makes using word boundaries impossible. He wants to do this on his path variable, a colon delimited string of directory names. In this situation the word boundary is :/, which isn't matched by perl's \b. \W comes closer, but still won't match if what he is looking for is at the begining of the line. And \W\/usr\/bin\W/ will match both /usr/bin and /usr/bin/X11.

So, you end up doing something like
s/^/ /;tr /:/ /;if ( \W\/usr\/bin\W/ )
to get it to work that way.

Re: grep -E (with regular expression) \b

Bill Hassell — Sat, 13 Mar 2004 14:24:21 GMT

To answer two questions about 10.20: no, grep will never have the -w option, and no, you do not have Perl version 5 in 10.20, only Perl 4 which is fairly incompatile with Perl 5 coding. 10.20 has been a dead end since Dec 2001.