I wanted to create a script that executes the ls command with the provided arguments. Why isn't it working?
I tried running it with for example -R, then it echoes (I echoed the same thing for comparison):
ls "-R"
but says that it cannot access "-R": no such file or directory. But when I execute ls "-R" in the terminal, it works.
The script itself:
text='ls'
while [[ $# -gt 0 ]]
do
text+=' "'
text+=$1
text+='"'
shift
done
echo 'ls "-R"'
echo $text
$text
The wrong quoting is the problem. " does not quote (and, which is the problem here, does not get removed!) if it is contained in a variable. Exception: eval $text
You should assign the parameters to an array and use ls "${vars[@]}" or ls "$@" instead.
Compare the two calls:
set -x
ls -l
ls "-l"
set +x
That's not how quoting work in bash. When entering
ls "-R"
quotes removal happens on the command, so the command becomes
ls -R
If, on the other hand, if you set
text='ls "-R"'
and run $text, the quote removal does not happen. See man bash under EXPANSIONS:
After the preceding expansions, all unquoted occurrences of the characters \, ', and " that did not result from one of the above expansions are removed.
" here resulted from parameter expansion, so it is not removed.
