Backreferences in sed returning wrong value

Question

I am trying to replace an expression using sed. The regex works in vim but not in sed. I'm replacing the last dash before the number with a slash so

/www/file-name-1 

should return

/www/file-name/1

I am using the following command but it keeps outputting /www/file-name/0 instead

sed 's/-[0-9]/\/\0/g' input.txt

What am I doing wrong?

Solution 1

You must surround between parentheses the data to reference it later, and sed begins to count in 1. To recover all the characters matched without the need of parentheses, it is used the & symbol.

sed 's/-\([0-9]\)/\/\1/g' input.txt

That yields:

/www/file-name/1

Solution 2

You need to capture using parenthesis before you can back reference (which start a \1). Try sed -r 's|(.*)-|\1/|':

$ sed -r 's|(.*)-|\1/|' <<< "/www/file-name-1" 
/www/file-name/1

You can use any delimiter with sed so / isn't the best choice when the substitution contains /. The -r option is for extended regexp so the parenthesis don't need to be escaped.

Comments

[user2052491]

if I use \1 it returns "\1 not defined in the RE"

[maljub01]

This is the correct answer. I didn't notice @user2052491 was not using the parentheses.

[Jack Wasey]

OS X / macOS has ancient UNIX tools. You can't use -r which means you need to escape the parentheses.

[phil294]

the backslashes before the braces are not necessary, are they?

[thenickdude]

OS X's sed uses -E instead of -r to enable extended regex