Bash variable scope

The problem is that processes put together with a pipe are executed in subshells (and therefore have their own environment). Whatever happens within the while does not affect anything outside of the pipe.

Your specific example can be solved by rewriting the pipe to

while ... do ... done <<< "$OUTPUT"

or perhaps

while ... do ... done < <(echo "$OUTPUT")

One more option:

#!/bin/bash
cat /some/file | while read line
do
  var="abc"
  echo $var | xsel -i -p  # redirect stdin to the X primary selection
done
var=$(xsel -o -p)  # redirect back to stdout
echo $var

EDIT: Here, xsel is a requirement (install it). Alternatively, you can use xclip: xclip -i -selection clipboard instead of xsel -i -p

Because you're piping into the while loop, a sub-shell is created to run the while loop.

Now this child process has its own copy of the environment and can't pass any variables back to its parent (as in any unix process).

Therefore you'll need to restructure so that you're not piping into the loop. Alternatively you could run in a function, for example, and echo the value you want returned from the sub-process.

http://tldp.org/LDP/abs/html/subshells.html#SUBSHELL

This should work as well (because echo and while are in same subshell):

#!/bin/bash
cat /tmp/randomFile | (while read line
do
    LINE="$LINE $line"
done && echo $LINE )