How do I use sed to store a variable? - bash

Question

I am trying to use sed to use as input for a variable. The user will choose from a list of files that have numbers before each to identify individual files. Then they choose a number corresponding to a name. I need to get the name of that file. My code is:
for entry in *; do
((i++))
echo "$i) $entry: "
done
echo What file # do you want to choose?:
read filenum
fileName=$(./myscript.sh | sed -n "${filenum}p")
echo $fileName ###this is to see if anything goes into fileName. nothing is ever output
echo What do you want to do with $fileName?
Ideally I would use () instead of the backtick but I can't seem to figure out how. I've looked at the links below, but can't get those ideas to work. I believe a problem may be that I am trying to include the filenum variable inside my sed.
https://www.linuxquestions.org/questions/linux-newbie-8/storing-output-of-sed-in-a-variable-in-shell-script-499997/
Store output of sed into a variable

Answer 1

Don't put backticks around $filenum. That will try to execute the contents of $filenum as a command. Put variables inside double quotes.
And if you do want to nest a backtick expression inside another set of backticks, you have to escape them. That's where $() becomes useful -- they nest without any hassle.
When you use sed -n, you need to use the p command to print the lines that you want to show in the output.
fileName=$(sed -n "${filenum}p" myscript.sh)
This will put the contents of line $filenum of myscript.sh in the variable.
If you actually wanted to execute myscript.sh and print the selected line of its output, you need to pipe to sed:
fileName=$(./myscript.sh | sed -n "${filenum}p")