Variable is not being echoed - linux

Question

When I run the following script in Bash 3.2.48:
#!/bin/bash
export var1='var1'
echo "UID=$UID"
if [ x"$UID" != x"0" ]
then
export var2='var2'
while ! { sudo -v; }; do { sudo -v; }; done;
sudo $0
exit
fi
echo $var1
echo $var2
exit 0
What I get as output is:
UID=1000
UID=0
var1
Why is var2 not exported and echoed? I'm pretty sure that the same script worked with older Bash versions.

Answer 1

you enter first time with UID == 1000, you enter the if clause
you sudo to execute the script with UID == 0;
sudo doesn't preserve the environment if env_reset is set in /etc/sudoers (default in most distros). You need sudo -E to preserve env.
you exit (before echoing)
from the sudo call you enter with clean env.
you enter with UID == 0
you don't enter the if clause, var2 is not set
you echo the variables.

Answer 2

The answer is much more simpler than is seems: you never echo those vars (when not running as root, obviously), because you already exit :))
Try avoiding/minimizing confusion by adding more appropriate/concise debug statements. For instance, use a single echo that contains everything that's relevant to your problem (i.e. process ID, user ID, var1, var2):
#!/bin/bash
export var1='var1'
if [ "$UID" != "0" ] ; then
export var2='var2'
while ! { sudo -v; }; do { sudo -v; }; done;
sudo $0
# this is "the key exit" ;-)
#exit
fi
echo "pid=[$$] uid=[$UID] var1=[$var1] var2=[$var2]"
With the exit commented out you get what you expect (obviously, in the "parent" process, as the "child" one - the one running as "root" - never reaches that part of the code that exports var2):
pid=[12346] uid=[0] var1=[var1] var2=[]
pid=[12345] uid=[1] var1=[var1] var2=[var2]
++ sometimes running scripts in debug mode (bash -x) helps too ;-)