Ruby replace string with captured regex pattern

Question

I am having trouble translating this into Ruby.

Here is a piece of JavaScript that does exactly what I want to do:

function get_code(str){
    return str.replace(/^(Z_.*): .*/,"$1");
}

I have tried gsub, sub, and replace but none seem to do what I am expecting.

Here are examples of things I have tried:

"Z_sdsd: sdsd".gsub(/^(Z_.*): .*/) { |capture| capture }
"Z_sdsd: sdsd".gsub(/^(Z_.*): .*/, "$1")
"Z_sdsd: sdsd".gsub(/^(Z_.*): .*/, "#{$1}")
"Z_sdsd: sdsd".gsub(/^(Z_.*): .*/, "\1")
"Z_sdsd: sdsd".gsub(/(.).*/) { |capture| capture }

Jake Berger · Accepted Answer · 2015-02-12 13:10:32Z

226

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"

edited Feb 12, 2015 at 13:10

Jake Berger

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answered Mar 27, 2012 at 22:14

Michael Kohl

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7 Comments

Vicky Chijwani Over a year ago

Note that this only works if the replacement string is inside single quotes. I wasted 5 minutes figuring that out.

Josh M. Over a year ago

@MarkThomas - often times we try the top/accepted answer first without reading the entirety of answers. That seems to generally be the most efficient means of fixing a problem. Give Vicky a break! :)

Jonathan Komar Over a year ago

@VickyChijwani Good comment, but also note that when using Ruby inline (on the command line with -e), it is more likely to see double quotes: printf "Punkinhead the name" | ruby -ne 'puts gsub /.*(the name)/, "Jonathans \\1"' because expression provided to -e is usually wrapped in single quotes.

Iulian Onofrei Over a year ago

@JagdeepSingh, It, by default, replaces all the occurrences.

Sandro L Over a year ago

@VickyChijwani When youn want to use double quotes you can wirte "\\1" instead of '\1'. Then something like "Z_123_2018".gsub(/\A(Z_\d+_)(\d{4})\z/, "\\1#{$2.to_i+1}") → "Z_123_2019" is possible.

|

Mark Thomas · Accepted Answer · 2013-07-18 15:26:15Z

40

\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.

edited Jul 18, 2013 at 15:26

answered Mar 27, 2012 at 22:20

Mark Thomas

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Comments

gaurav.singharoy · Accepted Answer · 2014-02-07 23:12:26Z

21

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

answered Feb 7, 2014 at 23:12

gaurav.singharoy

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2 Comments

vreen Over a year ago

didn't know I can do that. Nice!

RaphaMex Over a year ago

That block is very handy!

grumpit · Accepted Answer · 2014-10-20 17:10:09Z

5

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"

answered Oct 20, 2014 at 17:10

grumpit

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Comments

maerics · Accepted Answer · 2012-03-27 22:14:23Z

2

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

answered Mar 27, 2012 at 22:14

maerics

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Comments

Lisapple · Accepted Answer · 2019-11-09 00:36:59Z

1

$ 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)

answered Nov 9, 2019 at 0:36

Lisapple

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