how can a single command take stdin in one program and filename in other

Question

What i'm trying to do is pass argument into the second program in such a way that it takes cmdline-args and is opened with sudo but doesn't have anything to with stdin that i pass to sudo.

Below is an example to understand what i'm trying to do.

echo "mypass" | sudo -S nvim "filename"

Here sudo must take stdin and nvim must take a file-name from cmdline-arg. But nvim opens an unnamed file and puts stdin into that file and doesn't even care about the args i pass to it.

My end goal is not to only open file without entering password but to understand how can i pass arguments the way i want to

Is it even possible?

Answer 1

Simply pass the filename as first argument of your script:

echo "<your password>" | sudo -S nvim "$1"

and call the script with:

./script.sh myfile.txt

Note that the -S option of sudo behaves differently depending if the credentials are already cached or not. Better also use the -k option to invalidate the credentials:

echo "<your password>" | sudo -k -S nvim "$1"

Answer 2

Exactly as you have it seems like it ought to work.
Maybe try a bash call explicitly. That works even if you need a separate input redirection.

$ echo "$pw"|sudo -S bash -c "head -3 < foo"
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If you need to pass a file argument, maybe explicitly bind your stdin elsewhere.

$ echo "$pw"|sudo -S bash -c "head -5 foo </dev/null"
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Answer 3

Since the sudo password doesn't need to be repeated for two consecutive sudo commands, I should issue two sequential commands:

echo "mypass" | sudo -S -v 
sudo nvim "filename"         # will not prompt for password this time

Explanation: sudo -v just updates the credentials of sudo without executing any command.