Bash Sudo – How to Use Sudo Within a Function

bashfunctionsudo

I have written a function which acts in a similar way to tee but also pre-pends a datestamp. everything works fine except when i want to output to a file which is only root writable (in my case a logfile within /var/log). I've simplified the following code snippet to just include the bits which are not working:

#!/bin/bash
#script ~/test_logger.sh
logfile=/var/log/test.log
logger()
{
    while read data
    do
        echo $data >> $logfile
    done
    return 0
}
sudo ls ~ | logger

it works fine if i run the whole script like so sudo ~/test_logger.sh but i can't always do this since i want to use the logger function in files like ~/.bash_logout which are run automatically. i've tried putting sudo in front of the echo in the while loop but this does not work. any ideas?

Best Answer

it's generally bad practice to put sudo in a script. A better choice would be to call the script with sudo from ~/.bash_logout or wherever else you want to use it, if you must, or better still just make /var/log/test.log world-writable.

Related Solutions

Bash – Run nvm (bash function) via sudo

The problem here, as is so often the case, is about the different types of shell:

When you open a terminal emulator (gnome-terminal for example), you are executing what is known as an interactive, non-login shell.
When you log into your machine from the command line, or run a command such as su username, or sudo -u username, you are running an interactive login shell.

So, depending on what type of shell you have started, a different set of startup files are read. From man bash:

   When  bash is invoked as an interactive login shell, or as a non-inter‐
   active shell with the --login option, it first reads and executes  com‐
   mands  from  the file /etc/profile, if that file exists.  After reading
   that file, it looks for ~/.bash_profile, ~/.bash_login, and ~/.profile,
   in  that order, and reads and executes commands from the first one that
   exists and is readable.  The --noprofile option may be  used  when  the
   shell is started to inhibit this behavior.

In other words, ~/.bashrc is ignored by login shells. Since you are using the -i option to sudo, the startup files for the user's login shell are being read (from man sudo):

 -i, --login
             Run the shell specified by the target user's password data‐
             base entry as a login shell.  This means that login-specific
             resource files such as .profile or .login will be read by the
             shell.

So, what you can do is

Define the function in the user's ~/.profile or ~/.bash_profile instead. Bear in mind that ~/.profile is ignored if ~/.bash_profile exists. Also keep in mind that ~/.bash_profile is bash-specific so I would use .profile instead, just make sure that ~/.bash_profile does not exist.
Source ~/.nvm/nvm.sh from ~/.profile.

Bash – Executing a Bash Script Function with Sudo

I will admit that there's no simple, intuitive way to do this, and this is a bit hackey. But, you can do it like this:

function hello()
{
    echo "Hello!"
}

# Test that it works.
hello

FUNC=$(declare -f hello)
sudo bash -c "$FUNC; hello"

Or more simply:

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

It works for me:

$ bash --version
GNU bash, version 4.3.42(1)-release (x86_64-apple-darwin14.5.0)
$ hello
Hello!
$
$ FUNC=$(declare -f hello)
$ sudo bash -c "$FUNC; hello"
Hello!

Basically, declare -f will return the contents of the function, which you then pass to bash -c inline.

If you want to export all functions from the outer instance of bash, change FUNC=$(declare -f hello) to FUNC=$(declare -f).

Edit

To address the comments about quoting, see this example:

$ hello()
> {
> echo "This 'is a' test."
> }
$ declare -f hello
hello ()
{
    echo "This 'is a' test."
}
$ FUNC=$(declare -f hello)
$ sudo bash -c "$FUNC; hello"
Password:
This 'is a' test.

Best Answer

Related Solutions

Bash – Run nvm (bash function) via sudo

Bash – Executing a Bash Script Function with Sudo

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