Reomve elements from a path name

Shell Programming and Scripting
1

I have a srcipt that I pass a path into that will consist of 5 or 6 elements (/1/2/3/4/5/6). I want to always pull the third element out of the path name. How can i do this?

Thanks

2 
OLDIFS=$IFS; IFS=/; set -- $path; IFS=$OLDIFS
echo "$4"  # because the first element $1 will be empty
3

# echo "$1" | sed 's/\// /g' | awk '{print $3}'

4

Actually, if by "pull out" you mean "remove" rather than "extract" (as per the topic title), do as above, then glue back together all except the one you don't want. Something like

p=
n=
s=
while true
do
  case $# in 0) break;; esac
  i=$1
  shift
  # skip $4
  n=x$n
  case $n in xxxx) continue;; esac
  p="$p$s$i"
  s=/
done
echo "$p"

That got messier than I thought, but it uses no external programs.