i need help with a shell script that passes a variable. why does this not work?: echo $`echo 1` i want this to produce the same result as echo $1 but instead of the first option given at the commandline i get "$1" as result. here is the script if u are interested in what i want to achieve: any thoughts appreciated. #!/bin/bash if [ ! $ORACLE_HOME ]; then echo "ORACLE_HOME unset. exiting." exit 1 fi if [ $(basename `echo $ORACLE_HOME`) == "9i" ]; then LOGS="$ORACLE_HOME/j2ee/Oracle9iDS/log/ $ORACLE_HOME/j2ee/Oracle9iDS/application-deployments/forms/" fi if [ $(basename `echo $ORACLE_HOME`) == "9iAS" ]; then LOGS="$ORACLE_HOME/j2ee/Oracle9iAS/log/ $ORACLE_HOME/j2ee/Oracle9iAS/application-deployments/forms/" fi echo -e "\n\nLogfile sizes:\n\n" for l in $LOGS; do echo -e "\n$l\n" ls -lah $l done echo -e "\n\nGo to...\n\n" for i in $LOGS; do if [ ! $X ]; then X=1 else X=`expr $X + 1` fi echo "$X $i" done read -p "Choice: " SEL set $LOGS DIR=$`echo $SEL` echo $DIR #cd $`echo $SEL` && bash

Perhaps I'm missing a point but does DIR=$SEL not work for you? Otherwise I think eval() would help you.

> Perhaps I'm missing a point but does > DIR=$SEL > not work for you? Otherwise I think eval() would help you. well, SEL will hold a number...for example "1" in this case i want DIR to hold "$1" so i would like $$SEL. get it? but if i do it like that bash interprets it as $$ and SEL instead of $ and $SEL. :( eval() looks much like a c function to me is it? i would like to stick with bash. i know there is a bash builtin called eval but this does something completely different, doesnt it?