Please explain to me why the very last
echo statement is blank? I expect that
XCODE is 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=`expr $XCODE + 1`
and it too won't print outside of the while statement. I think I'm missing something about variable scope here, but the ol' man page isn't showing it to me.
Because you're piping into the while loop, a sub-shell is created to run the while loop.
Now this child process has its own copy of the environment and can't pass any variables back to its parent (as in any unix process).
Therefore you'll need to restructure so that you're not piping into the loop.
Alternatively you could run in a function, for example, and
echo the value you
want returned from the sub-process.
The problem is that processes put together with a pipe are executed in subshells (and therefore have their own environment). Whatever happens within the
while does not affect anything outside of the pipe.
Your specific example can be solved by rewriting the pipe to
while ... do ... done <<< "$OUTPUT"
while ... do ... done < <(echo "$OUTPUT")