How can I escape the quotes for the parameter of shell script?

Question

I have a shell script and want to be able to pass commands like: bash -c "COMMAND ARGS"

set -x
cmd='bash -c "ls /lib"'
docker run --rm ubuntu:bionic $cmd

But I get

+ cmd='bash -c "ls /lib"'
+ docker run --rm ubuntu:bionic bash -c '"ls' '/lib"'
/lib": -c: line 0: unexpected EOF while looking for matching `"'
/lib": -c: line 1: syntax error: unexpected end of file

How to escape these extra quotes?

and what about if I need to get the argument from the command line, using for example ./script.sh "bash -c 'ls lib'"

set -x
cmd=$1
docker run --rm ubuntu:bionic $cmd

I don't know why you're calling `ls` on startup, but if you're looking for a file, you should be using `find` instead. — Andy Lester, Apr 20 '22 at 15:31
Why do you need Bash at all here? `docker run image ls /lib` — tripleee, Apr 21 '22 at 07:45

score 2 · Answer 1 · answered Apr 20 '22 at 15:26

2

Use an array, not a regular parameter, to store the command.

set -x
cmd=(bash -c "ls /lib")
docker run --rm ubuntu:bionic "${cmd[@]}"

answered Apr 20 '22 at 15:26

chepner

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Any idea on how to get it working when passing the argument from CLI? – Odd Kviteberg Apr 21 '22 at 07:40
Why would you need the arguments to be in a variable at all? https://mywiki.wooledge.org/BashFAQ/050 – tripleee Apr 21 '22 at 07:44
I plan to make a script to run commands in a container. The commands themselves should be passed as arguments to the script – Odd Kviteberg Apr 21 '22 at 07:51

How can I escape the quotes for the parameter of shell script?

1 Answers1