Why escaped single quote doesn't work in grep?

Question

I am trying to match a string:

* * * * * /usr/local/bin/growlnotify -t 'helloTitle' -m 'helloMessage' -n 'myApp' -sw

with:

crontab -l | grep '^[^#].*/usr/local/bin/growlnotify.*\-n \'myApp\'.*$'

it doesn't work: nothing matched.

But:

crontab -l | grep '^[^#].*/usr/local/bin/growlnotify.*\-n.*$'

works very good:

* * * * * /usr/local/bin/growlnotify -t 'helloTitle' -m 'helloMessage' -n 'myApp' -sw

What is the problem with \'myApp\' ?

How to escape a single quote in grep/sed?

Solution

one can escape single quote with -e option and using hexadecimal representation:

echo -e '\047'
> '

or in grep using hexadecimal representation and -P option:

crontab -l | grep -P '^.*/usr/local/bin/growlnotify.*\-n \047myApp\047.*$'
> * * * * * /usr/local/bin/growlnotify -t 'helloTitle' -m 'helloMessage' -n 'myApp' -sw

ss64 as reference:

-P
   --perl-regexp
       Interpret PATTERN as a Perl regular expression.

Comments

[barlop]

-1 you really should've used a simpler example to demonstrate your point

[Scott - Слава Україні]

To clarify: "\" cannot be used to escape anything inside single quotes. This is not a grep/sed issue; this is a shell issue (bash, csh, etc…).

[adam]

If you are using GNU grep, you could have also replaced the single quotes ' enclosing the expression with double quotes " (2 chars modified) and removing the backslashes \ (3 chars removed). Copy/paste proof echo "* * * * * /usr/local/bin/growlnotify -t 'helloTitle' -m 'helloMessage' -n 'myApp' -sw" | grep "^[^#].*/usr/local/bin/growlnotify.*-n 'myApp'.*$"