UNIX for Dummies Questions & Answers

Hello,
I have a simple script such as

-----------------------------
#! /bin/sh


YEAR=`date -u +%Y`;
MONTH=`date -u +%m`;
DAY=`date -u +%d`;
DATE=$MONTH$DAY$YEAR
LOGFILES=auditTrail-$DATE
LOGMATCH=$LOGFILES\*
ARGUM=''

# find all files and write them to a file
find . -name "$LOGMATCH" > temp


cat ./temp | while read file_line
do
ARGUM=$file_line' '$ARGUM
echo $ARGUM
done

echo "final string contains"
echo $ARGUM
echo "nothing"

----------------------------

I was hoping that it will read a line one at the time and add to the variable ARGUM. It does exactly that within the loop. Yet, when it gets out the loop, last line
echo $ARGUM
return nothing

output looks like this
./log/auditTrail-02022007.10.out
./log/auditTrail-02022007.14.out ./log/auditTrail-02022007.10.out
./log/auditTrail-02022007.7.out ./log/auditTrail-02022007.14.out ./log/auditTrail-02022007.10.out
./log/auditTrail-02022007.1.out ./log/auditTrail-02022007.7.out ./log/auditTrail-02022007.14.out ./log/auditTrail-02022007.10.out
final string contains

nothing


-----------------------
I thought that variable ARGUM is global everywhere in the script.
What am I missing?

Thank you in advance

The while runs in a subshell.

So all changes made to ARGUM are gone when the loop is done
When you change it into"

while read file_line
do
ARGUM=$file_line' '$ARGUM
echo $ARGUM
done < cat ./temp

it will work fine

Typical behavior for Linux systems or the bash shell.

Thank you for the reply.
I tried, but now am getting

syntax error: `./temp' unexpected

remove the cat.

Indeed, the "cat" shouldn't be there, I copied/pasted a little too fast

thank you again. I did remove "cat"
and my while loop look as

while read file_line
do
ARGUM=$file_line' '$ARGUM
done < ./temp

echo $ARGUM

Yet, I am still getting emty output.
echo @ARGUM produces empty string.
Any other suggestings.
Thank you in advance

Switch ksh or bash. They won't do that to while loops.