In this test I'm expecting it to print "var1 is 999".
user@penguin:~$ for num in {1..3}; do export var$num=9999 ; echo var$num is $var$num ; done
var1 is 1
var2 is 2
var3 is 3
user@penguin:~$ echo $var1 $var2 $var3
9999 9999 9999
This prints the PID instead of the variabled named by the two variable names.
user@penguin:~$ for num in {1..3}; do export var$num=9999 ; echo var$num is $$var$num ; done
var1 is 316var1
var2 is 316var2
var3 is 316var3
In your output example you have $$var$num instead of $var$num which you have above. Seems like a typo but $$ is a special parameter that expands to the PID of the shell/subshell it's being executed from.
What you are looking for however is indirect parameter expansion:
for num in {1..3}; do
export "var$num"=9999
varname="var$num"
echo "var$num is ${!varname}"
done
If there is no requirement to export you could use an array like so:
declare -a array
for ((i=0;i<3;i++)); do
array[$i]=9999
echo "array[$i] = ${array[$i]}"
done
array[0]=9999 will set the array element index 0 to 9999
${array[0]} will reference the array element with index 0
etc.
