How to filter out Bash arguments

Ever want to know how to drop an argument (and value), --dir in this case, from a Bash script? Someone from my local LUG asked how to do it and this is what I came up with: Fun ;) #!/usr/bin/env bash args=("$@") myargs=() nextarg=-1 for ((i=0; i<$#; i++)) { if [ $nextarg == $i ]; then continue; fi case ${args[$i]} in --dir) nextarg=$((i+1)) ;; *) myargs+="${args[$i]} " esac } echo $myargs ./remove_dir.bash --dir foo --bar baz --bar baz

bash (2)