Read bash script arguments multiple times - bash

Question

I have a bash script which can be passed a number of different arguments and variables to be used by the script itself. Some of the parameters get assigned to variables. The second while loop does not appear to be running when the program tries to execute it. I've simplified the following script for confidentiality/simplicity reasons.
./myscript --dir2 /new/path
while :; do
case $1 in
--var1) $var1=$2
;;
--var2) $var2=$2
;;
"") break
esac
shift
done
$dir1=/default/directory
$dir2=/default/directory
while :; do
case $1 in
--dir1) $dir1=$2
;;
--dir2) $dir2=$2
;;
"") break
esac
shift
done
echo "Expect /default/directory, returned: $dir1"
echo "Expect /new/path, returned: $dir2"
Here's what my program would effectively return.
Expected /default/directory, returned: /default/directory
Expected /new/path, returned: /default/directory
Is there a better way to go about this? Or another way to iterate over the parameters originally passed to the script? Thanks for the help!

Answer 1

If you want to preserve your arguments, you can copy them into an array, and then restore the original list from that array later:
#!/usr/bin/env bash
# ^^^^ - must be invoked as bash, not sh, for array support
# Copy arguments into an array
original_args=( "$#" )
# ...consume them during parsing...
while :; do # ...parse your arguments here...
shift
done
# ...restore from the array...
set -- "${original_args[#]}"
# ...and now you can parse them again.

Answer 2

It's because your using shift which consumes the elements. Try using a for loop to iterate over the args. There's a few ways to do this. Here is my preferred method:
for elem in "$#" # "$#" has the elements in an array
do
... # can access the current element as $elem
done
Another way is to access them by index, you can look up a tutorial on bash array syntax for that.