What does \1 in sed do?

Question

I found this question really relevant to what I wanted: Parsing using awk or sed in Unix, however I can't figure out what the following does:

's/\([,=]\) /\1/g'

I know that g does a global substitution but really can't understand what's going on within the context of the question.

Answer 1

Here's a simple example:

$ echo 'abcabcabc' | sed 's/\(ab\)c/\1/' ababcabc $ echo 'abcabcabc' | sed 's/\(ab\)c/\1/g' ababab $ echo 'abcabcabc' | sed 's/\(ab\)\(c\)/\1d\2/g' abdcabdcabdc 

In the first command, only the first match is affected. In the second command, every match is affected. In both cases, the \1 refers to the characters captured by the escaped parentheses.

In the third command, two capture groups are specified. They are referred to by using \1 and \2. Up to nine capture groups can be used.

In addition to the g (global) operator (or without it, the first match), you can specify a particular match:

$ echo 'aaaaaa' | sed 's/a/A/4' aaaAaa 

Answer 2

\(...\) would capture the characters specified inside of the parens and \1 would be used to reference the first match, this is a part of regex.