Pattern match both suffix and prefix in shell script

Question

I have this variable, and I want to remove both prefix and suffix:

var="abcde"
echo "${${var#a}%e}" 

I tried this snippet in zsh, and it works properly - returns bcd. It uses pattern matching (prefix and suffix). Although, when I run it in posix compliant shell script (the reason I'm using pattern matching instead of bash string manipulation), I get output bad substitution.

What is the matter? Must I make another variable to store "${var#a}" into or can I do it in one line? I want posix compliant script.

Answer 1

No, you do not need to have another variable, and yes, you can do it in a single line.

This is probably the most portable solution but it will overwrite the variable value (still using a single variable as mentioned in the question). It uses only parameter expansion of the shell language:

var="abcde"
var="${var#a}" # remove prefix
echo "${var%e}" # remove suffix

One-liner may be always created as shown:

var="abcde"; var="${var#a}"; echo "${var%e}"

If you do not want to overwrite the variable, function arguments are perfect temporary variables. Although your example is trivial, you may create a function for it, which hides the extra variable used. This approach is especially useful since there are no local variables in the POSIX shell:

# this function runs in a subshell, so all new variables are lost on the return
remove_pre_post() (
    local_var="${1#a}" # uses the first argument
    echo "${local_var%e}"
)

var="abcde"
remove_pre_post "$var" # call function

Another POSIX-compliant solution may involve sed:

var="abcde"
echo "$var" | sed 's/^a//; s/e$//'

You provided only the example without any pattern matching. Keep in mind that once pattern matching is a thing, you may see one solution better than the other depending on the pattern complexity.