bash script command line passing double quotas does not work-CodePudding

this is my bash script, test.sh

#!/bin/bash

printf '1st printf:\n%s\n\n\n' 'find . -name "*.[ch]"'${1}

printf '2nd printf:\n%s\n' 'find . -name "*.[ch]"'" ${1}"

find . -name "*.[ch]"  ${1}

I run it as test.sh "-not -path \"*examples*\"" and here is the output:

1st printf:
find . -name "*.[ch]"-not


1st printf:
-path


1st printf:
"*examples*"


2nd printf:
find . -name "*.[ch]" -not -path "*examples*"

what I try to is to run find . -name "*.[ch]" -not -path "*examples*" inside the script. But the $1 is treated like a list or something? How do I construct the right command line inside the script?

CodePudding user response：

Use single quotes for static strings and double quotes for variables.

find . -name '*.[ch]' -not -path "$1"

Note: Escape *, to stop the shell from expanding it. Run your script this way:

./myscript '*examples*'

Or this way:

./myscript \*examples\*

CodePudding user response：

Don't put quotes (or escapes) in your data (i.e. parameters, variables, etc). Quotes go around data, not in data. If you want to pass multiple parameters to add to a find command, pass each one as a separate parameter (just as you would to find itself), and use "$@" to expand them (double-quotes required to avoid parsing weirdness).

#!/bin/bash
find . -name "*.[ch]" "$@"

Then run it as test.sh -not -path "*examples*"