What is the opposite of Regexp.escape?

Question

What is the opposite of Regexp.escape ?

> Regexp.escape('A & B')
=> "A\\ &\\ B"
> # do something, to get the next result: (something like Regexp.unescape(A\\ &\\ B))
=> "A & B"

How can I get the original value?

Answer 1

replaces = Hash.new { |hash,key| key } # simple trick to return key if there is no value in hash
replaces['t'] = "\t"
replaces['n'] = "\n"
replaces['r'] = "\r"
replaces['f'] = "\f"
replaces['v'] = "\v"

rx = Regexp.escape('A & B')
str = rx.gsub(/\\(.)/){ replaces[$1] }

Also make sure to #puts output in irb, because #inspect escapes characters by default.

Basically escaping/quoting looks for meta-characters, and prepends \ character (which has to be escaped for string interpretation in source code). But if we find any control character from list: \t, \n, \r, \f, \v, then quoting outputs \ character followed by this special character translated to ascii.

UPDATE:

My solution had problems with special characters (\n, \t ans so on), I updated it after investigating source code for rb_reg_quote method.

UPDATE 2:

replaces is hash, which converts escaped characters (thats why it is used in block attached to gsub) to unescaped ones. It is indexed by character without escape character (second character in sequence) and searches for unescaped value. The only defined values are control-characters, but there is also default_proc attached (block attached to Hash.new), which returns key if there is no value found in hash. So it works like this:

1. for "n" it returns "\n", the same for all other escaped control characters, because it is value associated with key
2. for "(" it returns "(", because there is no value associated with "(" key, hash calls #default_proc, which returns key itself

The only characters escaped by Regexp.escape are meta characters and control characters, so we don't have to worry about alphanumerics.

Take a look at http://ruby-doc.org/core-2.0.0/Hash.html#method-i-default_proc for documentation on #defoult_proc

Answer 2

using a regex replace using \\(?=([\\\*\+\?\|\{\[\(\)\^\$\.\#\ ]))\

should give you the string unescaped, you would only have to replace \r\n sequences with there CrLf counterparts.

"There\ is\ a\ \?\ after\ the\ \(white\)\ car\.\ \r\n\ it\ should\ be\ http://car\.com\?\r\n"

is unescaped to :

"There is a ? after the (white) car. \r\n it should be http://car.com?\r\n"

and removing the \r\n gives you :

There is a ? after the (white) car. 
 it should be http://car.com?

Answer 3

You can perhaps use something like this?

def unescape(s)
  eval %Q{"#{s}"}
end

puts unescape('A\\ &\\ B')

Credits to this question.

codepad demo

If you are okay with a regex solution, you can use this:

res = s.gsub(/\\(?!\\)|(\\)\\/, "\\1")

codepad demo

Answer 4

Try this

>> r = Regexp.escape("A & B (and * c [ e] + )")
# => "A\\ &\\ B\\ \\(and\\ \\*\\ c\\ \\[\\ e\\]\\ \\+\\ \\)"
>> r.gsub("\\(","(").gsub("\\)",")").gsub("\\[","[").gsub("\\]","]").gsub("\\{","{").gsub("\\}","}").gsub("\\.",".").gsub("\\?","?").gsub("\\+","+").gsub("\\*","*").gsub("\\ "," ")
# => "A & B (and * c [ e] + )"

Basically, these (, ), [, ], {, }, ., ?, +, * are the meta characters in regex. And also \ which is used as an escape character.

The chain of gsub() calls replace the escaped patterns with corresponding actual value.

I am sure there is a way to DRY this up.

Update: DRY version as suggested by user2503775

>> r.gsub("\\","")

Update:

following are the special characters in regex

    [,],{,},(,),|,-,*,.,\\,?,+,^,$,<space>,#,\t,\f,\v,\n,\r