Reomve elements from a path name - Shell Programming and Scripting

kgeasler · #1 (**permalink**) 03-26-2008

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

era · #2 (**permalink**) 03-26-2008

Code:

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

redhead · #3 (**permalink**) 03-26-2008

Quote:

Originally Posted by kgeasler

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

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

era · #4 (**permalink**) 03-26-2008

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

Code:

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.

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