Using parameter expansion as search and replace

Last modified: 2023-01-10

In this evening’s studies I came across this bash script in a tutorial by Rob Muhlenstein:

!#/bin/bash
echo -e ${PATH//:/\\n}

I could not make heads or tails of all these slashes and curly braces, since the output clearly indicated that search and replacement was being performed. I’m used to the sed / vim syntax: s/foo/bar

After some research I learned that ‘//’ is a global search and replace syntax of several text processing programs. It is known as parameter expansion in bash.

Example:

foo="1234567890"
echo "${foo//[0-9]/x}"

This replaces all the digits in the $foo variable with ‘x’, so the output would be xxxxxxxxxx

To do this with sed, you would do:

echo "$foo" | sed 's/[0-9]/x/g'

For more info:

man bash

/parameter expansion