Bash: Running a function as sudo on remote host?

Question

I have the following example script, to run commands on a remote host.

The script accepts the remote host's password from user input which we can use to login as that user and I expect also pass to sudo.

The problem is, I can't seem to get the runOnRemoteHostAsRoot function inside the second function to run using sudo, which I need.

How do I get this to work?

Platform at both ends is Ubuntu 18 or 20.

#!/bin/bash

read SSHPASS
export SSHPASS

runOnRemoteHost() {
    # ...
    whoami
    # ...
}
# ...
sshpass -e ssh -o PasswordAuthentication=yes "user@remotehost" "$(declare -f runOnRemoteHost); runOnRemoteHost" 2>&1
# ...


runOnRemoteHostAsRoot() {
    # ...
    whoami
    # ...
}
# ...
echo "${SSHPASS}" | sshpass -e ssh -o PasswordAuthentication=yes "user@remotehost" "$(declare -f runOnRemoteHostAsRoot); sudo --stdin runOnRemoteHostAsRoot" 2>&1
# ...

Expected output:

user
root

Actual output:

user
[sudo] password for user: sudo: runOnRemoteHostAsRoot: command not found

Answer 1

The suggestion from @Will is helpful in this instance, using sudo bash -c, then declaring and running the function:

sudo bash -c "$(declare -f runOnRemoteHostAsRoot); runOnRemoteHostAsRoot"

We'll use that line after passing the password through sshpass for passwordless login, like this:

echo '${SSHPASS}' | sudo --stdin bash -c '$(declare -f runOnRemoteHostAsRoot); runOnRemoteHostAsRoot'`

So using this in the above example:

#!/bin/bash

read SSHPASS
export SSHPASS


runOnRemoteHost() {
    # ...
    whoami
    # ...
}
# ...
sshpass -e ssh -o PasswordAuthentication=yes "user@remotehost" "$(declare -f runOnRemoteHost); runOnRemoteHost" 2>&1
# ...


runOnRemoteHostAsRoot() {
    # ...
    whoami
    # ...
}
# ...
sshpass -e ssh -o PasswordAuthentication=yes "user@remotehost" "echo '${SSHPASS}' | sudo --stdin bash -c '$(declare -f runOnRemoteHostAsRoot); runOnRemoteHostAsRoot'" 2>&1

Output:

user
root