How can I execute a bash function with sudo?

Question

I have a bash function defined in a global bashrc, which requires root privileges to work. How can I run it with sudo, e.g. sudo myfunction. By default it gives an error:

sudo: myfunction: command not found

Answer 1

Maybe you can do:

function meh() {
    sudo -v
    sudo cat /etc/shadow
}

This should work and saves you from typing sudo on the commandline.

Answer 2

#!/bin/bash

function smth() {
    echo "{{"
    whoami
    echo "}}"
}

if [ $(whoami) != "root" ]; then
    whoami
    echo "i'm not root"
    sudo $0
else
    smth
fi

Answer 3

You can export your function to make it available to a bash -c subshell or scripts that you want to use it in.

your_function () { echo 'Hello, World'; }
export -f your_function
bash -c 'your_function'

Edit

This works for direct subshells, but apparently sudo doesn't forward functions (only variables). Even using various combinations of setenv, env_keep and negating env_reset don't seem to help.

Edit 2

However, it appears that su does support exported functions.

your_function () { echo 'Hello, World'; }
export -f your_function
su -c 'your_function'

Answer 4

I would execute a new shell by having sudo execute the shell itself, then the function will run with root privileges. For example something like:

vim myFunction
#The following three lines go in myFunction file
function mywho {
    sudo whoami
}

sudo bash -c '. /home/kbrandt/myFunction; mywho'
root

You could even then go to make an alias for the sudo bash line as well.

Answer 5

sudo bash

then run your command