Unix shell: Finding the run-time directory of a script

I've been trying to come up with an elegant way, in a shell script, to find the full path of the directory in which the script is located.

So far, I have this, which works for bash but not for sh (sh does not support substring expansion on variables):

 #!/usr/local/bin/bash -x
 # Finds the full path to the directory containing this script
 
 dir_containing_this_script=""
 script_dirname=`dirname $0`
 first_char="${script_dirname:0:1}" ; # Starting at index 0, 1 character
 case "$first_char" in
 ("/")
     # It's a full path. Use it unmodified.
     dir_containing_this_script="${script_dirname}"
 ;;
 (*)
     # Strip leading . characters
     while [ "${first_char}" = "." ]; do
         script_dirname="${script_dirname:1}" ; # substring from index 1 to end
         first_char="${script_dirname:0:1}"
     done
     current_dir=`pwd`
     if [ -s "${script_dirname}" ] ; then
         dir_containing_this_script="${current_dir}/${script_dirname}"
     else 
         dir_containing_this_script="${current_dir}"
     fi
 ;;
 esac
 
 echo "$dir_containing_this_script"

Update: Eric M on The WELL posted this most excellent solution:

 dn=`dirname $0`
 path=`(cd $dn; pwd)`
 echo "path = $path"

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