How can I escape an arbitrary string for use as a command line argument in Bash?

Question

I have a list of strings and I want to pass those strings as arguments in a single Bash command line call. For simple alphanumeric strings it suffices to just pass them verb

Answer 1

Bash interprets exclamation marks only in interactive mode.

You can prevent this by doing:

set +o histexpand

Inside double quotes you must escape dollar signs, double quotes, backslashes and I would say that's all.

Answer 2

Whenever you see you don't get the desired output, use the following method:

"""\special character"""

where special character may include ! " * ^ % $ # @ ....

For instance, if you want to create a bash generating another bash file in which there is a string and you want to assign a value to that, you can have the following sample scenario:

Area="(1250,600),(1400,750)"
printf "SubArea="""\""""${Area}"""\""""\n" > test.sh
printf "echo """\$"""{SubArea}" >> test.sh

Then test.sh file will have the following code:

SubArea="(1250,600),(1400,750)"
echo ${SubArea}

As a reminder to have newline \n, we should use printf.