How can I properly escape a string to be passed to a shell with a command argument

Question

How can I properly escape arbitrary commands?

For example:

sudo -u chris sh -c 'echo "\"leftright\""'

The above echos:

"leftright"

How would I echo out:

"left'right"

I've tried the following which I would expect to work but does not:

sudo -u chris sh -c 'echo "\"left\'right\""'

I can't quite get my head round how it is parsed.

score 0 · Answer 1 · answered Jul 19 '15 at 19:29

0

sudo -u chris sh -c 'echo "\"left'\''right\""'

answered Jul 19 '15 at 19:29

neuron

Can you explain this? What other characters do I need to potentially escape to allow any arbitrary commands? – Chris Stryczynski Jul 19 '15 at 19:34
It's basically how you handle the opening & the closing quote. For your above example you could also write it as : `sh -c 'echo \"left"'\''"right\"'`. Check [this](http://wiki.bash-hackers.org/syntax/quoting#ansi_c_like_strings) for reference. Also check [here](http://stackoverflow.com/questions/1250079/how-to-escape-single-quotes-within-single-quoted-strings) for a good explanation by liori – neuron Jul 19 '15 at 19:47

Petr Skocik · Answer 2 · 2015-07-19T20:07:05.870

 sh -c 'echo "\"left'"'"'right\""'

OR

 sh -c 'echo "\"left'\''right\""'

will work.

Either of the above will pass

 echo "\"left'right\""

after -c.

When you're inside a single quoted string there is no escaping other than closing that string.

To pass a single quote, you can either use

"'"

or

\'

.

The shell will concatenate adjacent arguments that don't have an input field separator (usually space) in between.

2 Answers2