Ruby grep with string argument

Question

I want to use grep with a string as regex pattern. How can i do that?

Example:

myArray.grep(/asd/i) #Works perfectly.

But i want to prepare my statement first

searchString = '/asd/i'
myArray.grep(searchString) # Fails

How can i achieve this? I need a string prepared because this is going into a search algorithm and query is going to change on every request. Thanks.

Solution 1

Regular expressions support interpolation, just like strings:

var = "hello"
re = /#{var}/i
p re #=> /hello/i

Solution 2

Having something in quotes is not the same as the thing itself. /a/i is Regexp, "/a/i" is string. To construct a Regexp from string do this:

r = Regexp.new str
myArray.grep(r)

Solution 3

Edit:

I found a better way

myString = "Hello World!"
puts "Hello World!"
puts "Enter word to search"
input_search = gets.chomp
puts "Search insensitive? y/n"

answer = gets.chomp

if answer == "y"
  ignore = "Regexp::IGNORECASE"
else
  ignore = false
end

puts myString.lines.grep(Regexp.new(input_search, ignore))

My old Answer below:

Try making it a case or if statement passing if they want insensitive search on or off, etc

myString = "Hello World!"
puts "Enter word to search"
input_search = gets.chomp
puts "Search insensitive? y/n"
case gets.chomp
  when "y"
    puts myString.lines.grep(Regexp.new(/#{input_search}/i))
  when "n"
    puts myString.lines.grep(Regexp.new(/#{input_search}/))
end

Solution 4

Try this:

searchString = /asd/i
myArray.grep(searchString)

View more solutions

Comments

[gkaykck]

i can't,i have a string comes from a user, how can i merge that string with /asd/i ?

[gkaykck]

Regexp.new('/asd/i') returns a regex like (?-mix:\/asd\/i)?

[Ivaylo Strandjev]

@gkaykck the result is not the same on my machine, still what I wanted to stress on is that a regexp is not a string. I am simply showing you how to construct a regexp from a string. In the example above if str = "/u/" then r will become /\/u\// which is not what you want, I believe you can figure out how to do what you want

[Matstar]

Note that you may want to escape the value before you interpolate it, especially if it's provided by an untrusted user: /#{Regexp.escape(user_provided_string)}/i or Regexp.new(Regexp.escape(user_provided_string), "i"). This would for example treat a "." as a literal "." character and not as "any character". As a short but cryptic way to both escape a string and turn it into a regex, you can do Regexp.union(user_provided_string).