Unable to array values outside of function in shell script [duplicate] - linux

Question

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 method
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.

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 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

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

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.

Answer 6

#!/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 7

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)