When matching patterns with / in them, having to escape the / quickly becomes unwieldy, ugly:


With sed, perl or Vim, I would use a different delimiter for the regex, say ::

sed '\:/usr/share/man/: do something'
perl -ne 'print if m:/usr/share/man/:'
:g:/usr/share/man/: do something " Vim

How can I avoid this awk? Something like:

awk ':/usr/share/man/: {do something}'

The best I can think of is to use a variable:

awk -v pat='/usr/share/man/' '$0 ~ pat {do something}'

But that is very verbose compared to the sed/perl/vim method.

Of course, there might be other ways to match paths like /usr/share/man/, but that's not the only place where / could appear in a pattern.

  • 2
    you could do '$0 ~ "/usr/share/man/" {do something}' instead of using variable...
    – Sundeep
    Apr 19, 2017 at 3:31
  • 1
    @Sundeep I could, but the reason I brought it out is that this way I get to retain a layer of quoting (compare -v pat='"/' vs '$0 ~ "\"/"').
    – muru
    Apr 19, 2017 at 3:43
  • yup, that is true... and there is whole lot of escaping trouble while assigning something to awk variable which includes bash variables.. for ex: -v regex="^\\\s*$1\\\>" ... I don't think there is a way to use different regex delimiter than /regex/ so will have to use whatever fits case by case basis
    – Sundeep
    Apr 19, 2017 at 3:50

2 Answers 2


Just change /testregexp/ with $0 ~ "testregexp"

Simple exemple:

$ echo "a/b/c" | awk ' ( $0 ~ "a/b/c" ) { print "we have a winner" ; }'
we have a winner

another exemple

regexp='some regexp "with /lots/ of double quotes" "everywhere"'
awk -v reg="${regexp}" ' ( $0 ~ reg ) { action here... }'
  • See comments on the question.
    – muru
    Apr 19, 2017 at 16:05

Awk by itself doesn't have any easy, straightforward antidote for this. In the general case, switching to Perl (or down to sed for simple tasks) is the common workaround.

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