Bash variable scope

Question

Please explain to me why the very last "echo" statement is blank? I expect that it was incremented in the while loop to a value of 1:

#!/bin/bash
OUTPUT="name1 ip ip status" # normally output of another command with multi line output

if [ -z "$OUTPUT" ]
then
        echo "Status WARN: No messages from SMcli"
        exit $STATE_WARNING
else
        echo "$OUTPUT"|while read NAME IP1 IP2 STATUS
        do
                if [ "$STATUS" != "Optimal" ]
                then
                        echo "CRIT: $NAME - $STATUS"
                        echo $((++XCODE))
                else
                        echo "OK: $NAME - $STATUS"
                fi
        done
fi

echo $XCODE

I've tried using the following statement instead of the ++XCODE method

XCODE=`expr $XCODE + 1`

and it too wont print outside of the while statement. I think I'm missing something about variable scope here but the ol' man page isnt showing it to me.

Answer 1

Because you're piping into the while loop, a sub shell is created to run the while loop. Now this child process has it's 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

Answer 2

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")

Answer 3

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 )

Answer 4

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

Answer 5

 #!/bin/bash
 OUTPUT="name1 ip ip status"
+export XCODE=0;
 if [ -z "$OUTPUT" ]
----

                     echo "CRIT: $NAME - $STATUS"
-                    echo $((++XCODE))
+                    export XCODE=$(( $XCODE + 1 ))
             else

echo $XCODE

see if those changes help

Answer 6

Another option is to output the results into a file from the subshell and then read it in the parent shell. something like

#!/bin/bash
EXPORTFILE=/tmp/exportfile${RANDOM}
cat /tmp/randomFile | while read line
do
    LINE="$LINE $line"
    echo $LINE > $EXPORTFILE
done
LINE=$(cat $EXPORTFILE)

Answer 7

I got around this when I was making my own little du:

ls -l | sed '/total/d ; s/  */\t/g' | cut -f 5 | 
( SUM=0; while read SIZE; do SUM=$(($SUM+$SIZE)); done; echo "$(($SUM/1024/1024/1024))GB" )

The point is that I make a subshell with ( ) containing my SUM variable and the while, but I pipe into the whole ( ) instead of into the while itself, which avoids the gotcha.