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calling a aliased variable

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# 1  
Old 07-15-2002
calling a aliased variable
Issue:

i have variable A which is an alias for variable B which is equal to "THIS IS A TEST"

when every i echo variable A i only get the alias name for variable B, NOT the contents of variable B.



Code:
HOSTNAME# echo $TESTIT
+ echo THIS IS A TEST
THIS IS A TEST

HOSTNAME# ls -l
total 16
-rw-r--r--   1 user group        6 Jul 15 13:58 TEST.IT


HOSTNAME# for i in `ls`; do
DOH=`echo $i|awk -F. '{ print "$"$1$2 }'`
echo $DOH
done
$TESTIT


HOSTNAME#

Last edited by Optimus_P; 07-15-2002 at 05:55 PM..
# 2  
Old 07-15-2002
From your first statement:
Quote:
i have variable A which is an alias for variable B which is equal to "THIS IS A TEST"
I set A to equal value of B - echo $A - get nothing (B was nothing)
Set B and then A again - and the value shows up.
$ A=$B
$ B="This is a test"
$ echo $A

$ A=$B
$ echo $A
This is a test


As far as the code:
I don't see why you get two $TESTIT at the end. When I ran it, it only writes it out once ($TESTIT). I believe you are attempting to cat the file (show the contents of the file) Could you explain more of what you are attempting to do? What shell you are using and OS+version?
# 3  
Old 07-15-2002
sorry mate just got back from lunch.
OS: SOL 8
Shell: ksh

I have a list of files in a directory.

ie:
test.it
test.me
test.u

it takes this list and strips off the "." so i get testit, testme, testu.

these are actually variables that are sourced in from another file.
(i am sorceing it correctly)

so if you do a echo on those variables you get soemthing like the following:

echo $testit
THIS IS TEST IT

echo $testme
THIS IS TEST ME

echo $testu
THIS IS TEST U

Code:
. ./source-file
for i in `ls`; do
DOH=`echo $i|awk -F. '{ print "$"$1$2 }'`
echo $DOH
done

this only prints out the word $testit/$testme/$testu not the contents of what the variable is supposed to be.

now if by hand i say

doh=$testme
echo $doh
(it will then print the correct output) but useing the awk statement to fill the contents of $doh does not seem to work.
# 4  
Old 07-15-2002
just so you know what the actual script looks like.
Code:
. /var/adm/scripts/dfh

INDIR="/drop"
ARCHIVEDIR="/archive_messages/exported"
OUTDIR="/export/manual"
SENDDIR="/export/mbi"
FUUSER="user"
FUGROUP="group"
OPENFILE=`/usr/local/bin/lsof +d ${INDIR}|grep ${FUUSER}|awk -F"/" '{ print $4 }'`

set -x

if [[ -d ${INDIR} ]] ; then
        cd ${INDIR}
        for file in `ls` ; do

DFHNAME=$`echo $file | awk -F. '{ print $1 $2 }'`

                if [[ $file != ${OPENFILE} && `tail -1 $file|awk '{ print $1 }'` = "97" ]]; then
                        chown $FUUSER:$FUGROUP $file
                        echo "$DFHNAME" > /$OUTDIR/$file.tmp
                        cat /$INDIR/$file >> /$OUTDIR/$file.tmp
                        rm -f $file
                        cp -p  /$OUTDIR/$file.tmp $ARCHIVEDIR/$file.`date +%Y%m%d%H%M%S`
                        mv /$OUTDIR/$file.tmp $SENDDIR/$file.`date +%Y%m%d%H%M%S`
                                        fi
        done
fi

# 5  
Old 07-15-2002
Quote:
now if by hand i say
doh=$testme
echo $doh
(it will then print the correct output) but useing the awk statement to fill the contents of $doh does not seem to work.
Realize you are not setting DOH to a value of another variable - you are setting it to "$" $1$2. This is two different things. Just as I stated earlier, if B is nothing setting A=$B give you nothing.

You are stating DOH to be equal to a string "$TESTIT" not a value of the variable $TESTIT. I'm not sure the work around (I've been looking...). Will get back to it tomorrow if no one else has (must go home as the EOD is here).
# 6  
Old 07-15-2002
If I'm understanding this correctly, try this:

# b="this is a test"
# a="b"
# eval echo \$$a
this is a test
# 7  
Old 07-16-2002
gsatch: i dont have the eval command. i will see if i can download it.

RTM: your help is much appricated. Yes, shortyly after i posted last night i went home also. and your answer makes complete sense; something i had not thought of.
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