Problem with regular expression in gawk (‘\<' not working)

gawkregular expression

I am trying to get into bash a little bit. I was going through this guide but the second example in this subsection on gawk doesn't seem to work.

The problem is this command:

ls -l | awk '/\<(a|x).*\.conf$/ { print $9 }'

It works only if I replace \< with a space. I also tried using \y, but no luck there either.

Does anyone have any idea what the problem might be here?

Thanks 🙂

Best Answer

The GNU awk manual (sec. 3.5) documents that the regex \< is gawk-specific and thus one should not expect it to work in other implementations.

According to man mawk, if you place a backslash in front of a nonspecial character, then the backslash is removed. Thus, under mawk, \< is interpreted simply as an angle bracket character.

Examples

I simplified the regex to provide examples of the different behavior:

$ echo -e " a\n ab.conf\n <ac.conf" | gawk '/\<(a|x)/ { print}'
 a
 ab.conf
 <ac.conf
$ echo -e " a\n ab.conf\n <ac.conf" | mawk '/\<(a|x)/ { print}'
 <ac.conf

Again, gawk interprets \< as the beginning of a word while mawk interprets it simply as an angle bracket.

What does POSIX say about this issue

The GNU awk manual explains:

If you place a backslash in a string constant before something that is not one of the characters previously listed, POSIX awk purposely leaves what happens as undefined.

In other words, in this case, the different awk interpreters are free to make their own decisions.

Related Solutions

Grep regular expression solution (greedy not working)

The pattern *: will match everything until the last :. To stop at the next : you need *?:. E.g.:

% grep -Poz '^:MENU1\K[\w\W]*?:' data.txt 

0. public
1. admin
2. webmail

:

You can strip the first line by matching the newline before your \K. E.g.:

% grep -Poz '^:MENU1\n\K[\w\W]*?:' data.txt 
0. public
1. admin
2. webmail

:

To eat the empty line and the : you can match and discard that text. E.g.:

% grep -Poz '^:MENU1\n\K[\w\W]*?(?=\n+:)' data.txt 
0. public
1. admin
2. webmail

next we can simplify your character class, to match on anything but ::

% grep -Poz '^:MENU1\n\K[^:]*?(?=\n+:)' data.txt 
0. public
1. admin
2. webmail

And finally we can rewrite the initial part of the match:

% grep -Poz '(?<=:MENU1\n)[^:]*?(?=\n+:)' data.txt 
0. public
1. admin
2. webmail

This is similar to what @terdon came up with, but this takes care of the blank lines without another call to grep.

This final regex makes use of look-around assertions. The (?<=pattern) is a look-behind assertion that lets you match the pattern but not include it as part of the output. The (?=pattern) is a look-ahead assertion and lets us match on the trailing pattern without including it in the output.

Awk exit code if regular expression did not match

You can set a variable to hold the return code, then negate the variable before quitting:

netstat -nap    |
grep "LISTEN\b" |
awk '$4 ~ /:80$/ {rc = 1; print $NF}; END { exit !rc }'

If you don't need \b, then you can remove grep part:

netstat -nap | awk '/LISTEN/ && $4 ~ /:80$/ {rc = 1; print $NF}; END { exit !rc }'