How can I run the sudo command in Python code under CentOS in Docker

Question

I am trying to access the docker image labels from Python as follows

hostname = socket.gethostname()
cmd = "sudo curl --unix-socket /var/run/docker.sock http:/containers/" + hostname + "/json"
output = os.popen(cmd).read()

But, the thing is I am getting the following error:

We trust you have received the usual lecture from the local System
Administrator. It usually boils down to these three things:

    #1) Respect the privacy of others.
    #2) Think before you type.
    #3) With great power comes great responsibility.

sudo: no tty present and no askpass program specified

It's one of the fancy messages by Unix from some other posts I read from StackOverflow.

I am following the below link

https://stackoverflow.com/questions/37439887/how-to-access-the-metadata-of-a-docker-container-from-a-script-running-inside-th

Only thing is I want to run these things from Python not from the Terminal. Also, FYI, I get the response when I run the same command from the terminal.

I tried appending the following piece in Dockerfile

RUN echo "root ALL=(ALL) NOPASSWD:ALL" >> /etc/sudoers

Thanks

Answer 1

sudo configuration by default has requiretty set so that password authentication is known as reliably as possible to be presented to an actual meatbag rather than to standard input. This is checked even if NOPASSWD is set for a specific sudoers entry.

First, I would suggest you look long and hard at why you need to sudo curl in the first place. This is highly ill-advised as a general rule -- you don't need to be 'root' to run curl to pull something from a remote host. Run curl as a non-priveleged user and if you need superuser access to do something with whatever you've fetched, do that separately.

All that said, you can either disable requretty for the specific sudoers entry or for the entire configuration, and a TTY will no longer be required.