grep -E (with regular expression) \b

[Reggie Chang]

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

[curt larson_1]

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

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

[Reggie Chang]

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

[Bill Hassell]

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.

[Reggie Chang]

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

[curt larson_1]

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

[Reggie Chang]

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

[curt larson_1]

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

[curt larson_1]

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

[Mark Grant]

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}"

[john korterman]

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.

[curt larson_1]

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.

[Bill Hassell]

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.