I need concise and readable function to be used in bash script which will take as input:

  • absolute path (e.g. /home/user/tmp/data/sample1.txt )
  • new filename name (e.g. extended_sample.dat )

and return

absolute path with new filename (e.g. /home/user/tmp/data/extended_sample.dat )

Thank you in advance!

3 Answers 3


There's 2+1(bonus) commands that make life easy with dealing with path names:

  • basename - strip directory and suffix from filenames
  • dirname - strip last component from file name

  • readlink - print resolved symbolic links or canonical file names

So to answer your question:


new_path="$(dirname $old_path)/$new_file"

Or to change the path name of the original file:


new_path="$alt_path/$(basename $old_path)"

Extract full location of the running script (works if the script is executed off its own file):

dirname $(readlink -f ${0})

With these 3 commands you can do most of your path manipulations easily.


Note, the return statement in bash functions is used to return a numeric value as a status code.

How it may look in most simple case:



function get_new_path() { echo "${1%/*}/$2"; }

new_path=$(get_new_path "$path" "$new_name")
echo "$new_path"

The above will output:

  • 2
    Can't you just do "$(dirname "$1")/$2" ?
    – xenoid
    Feb 18, 2018 at 12:41

Just playing around with bash namerefs:

path_replace_file () {
    local -n pathvar="$1"


printf 'pathname before = %s\n' "$pathname"
path_replace_file pathname extended_sample.dat
printf 'pathname after  = %s\n' "$pathname"


pathname before = /home/user/tmp/data/sample1.txt
pathname after  = /home/user/tmp/data/extended_sample.dat

The path_replace_file function takes a variable name as its first argument. In the function pathvar acts like an alias (a name reference) for that variable.

The function replaces the filename portion of the value of $pathval with whatever is given as the second argument to the function.

This requires bash 4.3 or later.

Changing local -n to typeset -n will make it work in ksh93 too.

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