Ruby replace string with captured regex pattern

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"

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\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.

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

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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"

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$ variables are only set to matches into the block:

"Z_sdsd: sdsd".gsub(/^(Z_.*): .*/) { "#{ $1.strip }" }

This is also the only way to call a method on the match. This will not change the match, only strip "\1" (leaving it unchanged):

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

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