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This question already has an answer here:

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.

marked as duplicate by Gilles command-line Jul 19 '15 at 22:04

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

  • 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 for reference. Also check here for a good explanation by liori – neuron Jul 19 '15 at 19:47
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 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.

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