sh temp1.sh Gold.txt Silver.txt
2
Gold.txt
$2
Silver is second to gold. It is a unique position in a competition.
cat: cannot open $2
tstetlx () /appl/edw/apps/scripts/scenario1> vi temp1.sh
i=$#
echo $i
echo $1
echo $`echo $i`
#cat "$`echo $i`"
cat $2
cat "\$$i"
The below command is not printing the contents of the second file passed as an argument to the file.
cat "\$$i"
Not clear to me what you are trying to do. Consider using $1 or $2.
The reason that line fails is because the parameter for cat is literally $2 - no attempt by the shell to substitute $2 as if it were a variable.
This is one way to handle the problem. Note that this is a bash solution.
i=$#
echo $i
echo $1
echo $`echo $i`
#cat "$`echo $i`"
cat $2
cat "\$$i"
declare -a arr=( "$0" $@ )
echo ${arr[i]}
cat ${arr[i]}
Make sure a file called Silver.txt exists in the same directory as temp1.sh.
As for the variable indirection you are trying to do with cat "\$$i", you want eval:
eval cat "\$$i"
However, eval can be dangerous. Make sure the variable you will be eval'ing is valid, especially if its contents comes from user input. Run this and see what I mean:
eval cat "\$$i;ps -ef"
It will run a command if it can be manipulated to contain one. See this post for a better discussion on why and alternatives: Why should eval be avoided in Bash, and what should I use instead?.
