Ruby replace string with captured regex pattern

Question

Try '\1' for the replacement (single quotes are important, otherwise you need to escape the \):

"foo".gsub(/(o+)/, '\1\1\1')
#=> "foooooo"

But since you only seem to be interested in the capture group, note that you can index a string with a regex:

"foo"[/oo/]
#=> "oo"
"Z_123: foobar"[/^Z_.*(?=:)/]
#=> "Z_123"

\1 in double quotes needs to be escaped. So you want either

"Z_sdsd: sdsd".gsub(/^(Z_.*): .*/, "\\1")

or

"Z_sdsd: sdsd".gsub(/^(Z_.*): .*/, '\1')

see the docs on gsub where it says "If it is a double-quoted string, both back-references must be preceded by an additional backslash."

That being said, if you just want the result of the match you can do:

"Z_sdsd: sdsd".scan(/^Z_.*(?=:)/)

or

"Z_sdsd: sdsd"[/^Z_.*(?=:)/]

Note that the (?=:) is a non-capturing group so that the : doesn't show up in your match.

 "foobar".gsub(/(o+)/){|s|s+'ball'}
 #=> "fooballbar"

If you need to use a regex to filter some results, and THEN use only the capture group, you can do the following:

str = "Leesburg, Virginia  20176"
state_regex = Regexp.new(/,\s*([A-Za-z]{2,})\s*\d{5,}/)
# looks for the comma, possible whitespace, captures alpha,
# looks for possible whitespace, looks for zip

> str[state_regex]
=> ", Virginia  20176"

> str[state_regex, 1] # use the capture group
=> "Virginia"

def get_code(str)
  str.sub(/^(Z_.*): .*/, '\1')
end
get_code('Z_foo: bar!') # => "Z_foo"