Export command giving Variable Name vs the Value set for the Variable

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1

I'm having an issue when I export within my program. I'm getting the variable name, not the variable value. I have a configuration file (config.txt) that has the values of the variables set as so:

set -a
export ARCHIVEPOSourceDir="/interfaces/po/log /interfaces/po/data"
export ARCHIVEPRSourceDir="/interfaces/pr/log /interfaces/pr/data"

Then I have a shell script that I'm passing the configuration file and the interface name to as follows:

./CleanUpFiles.sh -C config.txt -P ARCHIVEPO

and here's the code in CleanUpFile.sh:

#!/usr/bin/ksh
PROG_NAME=`basename $0`

Usage ()
{
 echo "Usage: $PROG_NAME -C ConfigFile(config.txt) -P ARCHIVE_Interface_Name" 
 exit 1
}

if [ $# -ne 4 ]
then
 Usage
fi

while getopts C:P: Opt
do
    case ${Opt} in
        C) INTERFACE_CONFIG_FILE=${OPTARG}
           ;;	
        P) ARCHIVE_Interface_Name=${OPTARG}
           ;;
        ?) Usage 
    esac
done

. ${INTERFACE_CONFIG_FILE} 

export SOURCEDIR=${ARCHIVE_Interface_Name}SourceDir
echo ${SOURCEDIR}

So I pass ARCHIVEPO as the value for ARCHIVE_Interface_Name parameter to CleanUpFiles.sh. When the program executes: export SOURCEDIR=${ARCHIVE_Interface_Name}SourceDir, my expectation was that it would give back the values, "/interfaces/po/log /interfaces/po/data", but instead I get back "ARCHIVEPOSourceDir". If I hard code the statement as:export SOURCEDIR=${ARCHIVEPOSourceDir}, then I correctly get back the values, "/interfaces/po/log /interfaces/po/data". I need to be able to pass it as a parameter though, because this code is going to be reused by different interfaces. Any suggestions on a way around this?

Thanks,
Les...

2

Try:
eval export SOURCEDIR=\$${ARCHIVE_Interface_Name}SourceDir

3

Thank ya Perderabo! That worked!