Sed capture group

Question

I have a problem with sed unable to correctly capture the \1 capture group in the below example:

echo '//apple' | sed "s/^\(\/\)\{0,2\}apple\(\/\)\?/\1banana\2/"
                         -------------              --
                            ^                        ^
                            |                        |__ writing back what it captured
                            |
                            |_ this is the capture group that I have problem capturing

I expected the output of: //banana correctly capturing the //. How can I make sed to correctly the \1 capture group?

Answer 1

You only captured one /.

You need to use

echo '//apple' | sed 's/^\(\/\{0,2\}\)apple\(\/\)\?/\1banana\2/' # POSIX BRE
echo '//apple' | sed -E 's~^(/{0,2})apple(/)?~\1banana\2~'       # POSIX ERE variant

See the online demo. The POSIX BRE pattern matches

• ^ - start of string
• \(\/\{0,2\}\) - Group 1: zero, one or two occurrences of a / char
• apple - a fixed string
• \(\/\)\? - an optional Group 2 matchong a / char.

Note you do not need to escape capturing parentheses, nor ? and {min,max} quantifiers when using POSIX ERE regex that is enabled with the -E option.

You need not escape / chars in the sed command when / is not used as a regex delimiter char.

Also, see this regex graph:

Answer 2

With an alternate delimiter and -E use cleaner and easier to read:

echo '//apple' | sed -E 's~^(/{0,2})apple(/{0,2})~\1banana\2~'