bash while loops and pipes

In bash, if you pipe into a while loop, the while loop is run in a subshell. This means that you're going to be very disappointed if you were hoping to capture data/variables within the loop.

The work around is to not have the pipe there - which is possible through process substitution.

e.g.

echo -e "one\ntwo\nthree" | \
 while read name;
  do val=$name;
  echo $val;
 done;
echo $val

This outputs:

one
two 
three

Notice that the final value "three" is not printed twice as 'val' no longer contains a value.

And the work around:

while read name; 
 do val=$name;
 echo $val;
done < <(echo -e "one\ntwo\nthree")
echo $val

which now outputs:

one
two
three
three

Success! (or, if you're a "Bill and Ted's Bogus Journey" fan - Station!).