How to preserv user id/previleges while root in a bash script.

[octius4u]

Hi everybody.

I've tried to google for answers first but without any luck, so if I missed an answer please excuse and redirect me.

My Situation
I'm writing an post-install tweak script in /bin/bash, and because I don't want to have 15 sudo's and
be asked for the password to many times etc. I'm running it as root

sudo bash MyPostTweaks.sh

The script looks basically like this:

#!/bin/bash
apt-apt-repository -y $ppa_one
apt-get update 
apt-get install -y $app_one

#Now some tweaks for the user
wget $WebAddress --directory-prefix=$HOME/Pictures
# or something like
cp ../myconfs/myapp2.conf $HOME/.config/myapp2/

Now my Problem

I have had problems where programs have worked badly, by not having permissions, or even worse even though the *.desktop was install and present /usr/share/applications clicking on it would not work. I began to notice whats wrong because it tried to run the program from the terminal and read some error messages. The same problem was not present if I stated the program from a second user account. I understand that the reason for the permissions/privileges issues is the script being ran as root. I also know about sudo chmod or sudo chown but since I don't know where the issue lies because I've even tried

sudo chmod -R 755 ~/* && sudo chown -R octius4u:octius4u ~/*

.

My Question

While in user octius4u i run the script as sudo bash MyPostTweaks.sh essentially putting me in root's account. If I run the script on a computer where the user name has changed, for example because I got bored of the old user name, how can I determine it?
In order to run some thing like

sudo -u $NormalUser cp ../myconfs/myapp2.conf $HOME/.config/myapp2/

Can I use or combine some of the following commands?

w
id -un
whoami
in order to get something similar to

if [ $w_user = $(id -un) ]; then; sudo -u $w_user cp $path/$file /$newpath

Thank you in advance for any help or insights you can provide.

[Lars Noodén]

You should be able to determine the original user's account with the environment variable SUDO_USER. That's the only one I can find that contains the original account, along with SUDO_GID and SUDO_UID.

[octius4u]

I think that would work great! Thank you very much.

Would you know how to determine the $SUDO_USER's home path?
I am asking because i know that in some cases the home directory is not just /home/octius4u but it could be /home/o/octius4u or something else.

[Lars Noodén]

The only way to be sure would be to poll /etc/passwd to find what is really set there:

h=$(awk -F ':' "\$1==\"$SUDO_USER\" { print \$6 }" /etc/passwd);
echo $h

Mind all the escapes. The shell doesn't make it easy to work with $ and " inside strings. The field separator is set to a colon so that /etc/passwd makes sense to the rest of awk.

Edit: be sure to check that $h is not empty before trying to use it in path to do something automatic..

[octius4u]

I have entered echo $h before running your code and it was empty and after your code it did contain the correct user path even if it was enter from within root.
Even if I am not (yet) familiar with awk I can use it by simple copy and pate (for now) and it seems to work just perfectly.

I consider my question answered and my problem solved.

You have help me a lot. Thank you very much.

[Lars Noodén]

Glad it saved time. awk is simple once the main concept is understood: it's a bunch of if-then statements, but the now traditional markers are left out. In pseudo code here is the same bit

set field_separator to ':'

read /etc/passwd through to the last line

split each line into fields separated by ':'

if the first field of a line equals the value "$SUDO_USER" then print the content of the sixth field

It's great for very short scripts or one-liners. Anything more and you probably need perl or python.

[octius4u]

If you don't mind, I have a related question.

Regarding installing new ppa's I wish to make sure that the ppa address isn't already installed.
Could I use awk verify if it already exists, or would there be a easier way?

[Lars Noodén]

grep might work as an even simpler test.

grep -rl '^deb http://ppa.launchpad.net/ubuntu-wine/ppa/ubuntu' /etc/apt/* 2>/dev/null

That's a search for an exact string at the beginning of the line.

The files to look in would be in /etc/apt/ or /etc/apt/sources.list.d/ and they would contain the pointer to the PPA or not, the -r makes the search recursive.
Directories can't be grepped so they make noise (warnings) so 2> redirects stderr to /dev/null

[octius4u]

I have looked up "/dev/null" here at askubuntu.com and here at stackoverflow.com
Would the following therefor be correct?

grep -rl '^deb http://ppa.launchpad.net/ubuntu-wine/ppa/ubuntu' /etc/apt/* 2>/dev/null


if [ $(/dev/null) = "0" ]; then
      # Adding new ppa's
else
      echo "it exists already"
fi

[Lars Noodén]

The variable to test after the fact would be $?

But you can run the test directly on grep itself.

if ! grep -rl '^deb http://ppa.launchpad.net/ubuntu-wine/ppa/ubuntu' /etc/apt/* 1>/dev/null 2>&1;  then
      # Adding new ppas
else
      echo "it exists already"
fi

To cut down on the noise in the script, 1 is stdout, 2>&1 sends stderr to stdout, which is getting sent to the bit bucket (/dev/null).