Creating a regexp from string without quoting

Question

My code will receive a parameter containing a string representation of a regular expression. It is probable that the strings would be like "/whatever/" with slashes at the beginning and end. Given a string,

str = "/^foo.*bar$/"

I would like to create a regular expression from that string.

When I do:

pat = Regexp.new(str)
# => /\/^foo.*bar$\//
pat.match "foolishrebar"
# => nil

all of the special characters are quoted. I have not figured out how not to quote the string.

When I create a pattern directly with /pattern/, it works fine.

pat = /^foo.*bar$/
pat.match "foolishrebar"
# => #<MatchData "foolishrebar">

As an aside, you almost always want \A and \z (beginning/ending of string) instead of ^ and $ (beginning/ending of line) in Ruby. — mu is too short
– mu is too short, Commented Jan 28, 2019 at 18:14
If you really mean what you are asking for, then you can do eval(str). But I am not posting this as an answer because I know that there are stupid people around who would downvote such an answer as soon as I post one. — sawa
– sawa, Commented Jan 29, 2019 at 4:38

Silvio Mayolo · Accepted Answer · 2019-01-28 17:45:47Z

4

When you use Regexp.new, don't start and end your string with /. Just let str = '^foo.*bar$'. The only things being escaped are the beginning and ending slashes; the metacharacters are fine.

answered Jan 28, 2019 at 17:45

Silvio Mayolo

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1 Comment

m0j0 Over a year ago

Also, I can use str.tr( '/', '') to remove any slashes from the string before passing it to Regexp.new.

Cary Swoveland · Accepted Answer · 2019-01-28 18:18:28Z

0

str = '^foo.*bar$'

r = /#{str}/
  #=> /^foo.*bar$/ 

"foolishly, the concrete guy didn't use rebar".match?(r)
  #=> true

answered Jan 28, 2019 at 18:18

Cary Swoveland

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