Quote a path in a shell script

Question

I have a shell script that uses base64 to encode a value and store it in a variable.

encoded="$(cat $pathtofile|base64 -w 0)"

This worked until I ended up with a $pathtofile that had a special character in it. Now I'm trying to figure out how to quote the $pathtofile so that cat gets the right file. When I do this

encoded="$(cat '$pathtofile'|base64 -w 0)"

I end up with an error because it doesn't expand $pathtofile, but prints it literally. I have tried several other combinations, but they all result in a misquoted path.

How can I end up with a quoted $pathtofile?

Possible duplicate of Difference between single and double quotes in Bash — Benjamin W.
– Benjamin W., Commented Jul 13, 2018 at 2:39

John Kugelman · Accepted Answer · 2022-09-15 02:22:40Z

3

Double quotes can be nested when you use $(...).

encoded="$(cat "$pathtofile" | base64 -w 0)"

For what it's worth, the outer set of quotes is optional. They're not needed in variable assignments. Remove them if you like.

encoded=$(cat "$pathtofile" | base64 -w 0)

Also, congratulations, you've won a Useless Use of Cat Award!

encoded=$(base64 -w 0 "$pathtofile")

edited Sep 15, 2022 at 2:22

answered Jul 13, 2018 at 0:28

John Kugelman

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1 Comment

Daniel Watrous Over a year ago

Thanks for the useless use of cat award ;)

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