在 Ruby 字符串中查找哈希值

Question

I have a string that looks like this:

greetings = <<-GREETINGS
[{a: "hello", b: "goodbye"}, {a: "guten tag", b: "auf wiedersehen"}]
GREETINGS

How can I split this string, so as to return an Array of the Hashes?

Answer 1

require 'json'

JSON.parse(greetings.gsub(/(\p{Alnum})+:/, '"\1":')).
     map { |h| h.transform_keys(&:to_sym) }
  #=> [{:a=>"hello", :b=>"goodbye"},
  #    {:a=>"guten tag", :b=>"auf wiedersehen"}] 

The steps are as follows:

str = greetings.gsub(/(\p{Alnum})+:/, '"\1":')
  #=> "[{\"a\": \"hello\", \"b\": \"goodbye\"}, {\"a\": \"guten tag\", \"b\": \"auf wiedersehen\"}]\n"

arr = JSON.parse(str)
  #=> [{"a"=>"hello", "b"=>"goodbye"},
  #    {"a"=>"guten tag", "b"=>"auf wiedersehen"}] 
arr.map { |h| h.transform_keys(&:to_sym) } 
  #=> [{:a=>"hello", :b=>"goodbye"},
  #    {:a=>"guten tag", :b=>"auf wiedersehen"}] 

The regular expression reads, "match one or more alphanumeric characters in capture group 1 followed by a colon". The \\1 in gsub 's replacement string is replaced with the contents of capture group 1.

See JSON::parse , String#gsub and Hash#transform_keys .

Answer 2

If you trust the source (and only if you trust the source, which is probably not true):

data = eval(str)

If you do not (which should be close to 100% of cases), then you can use the parsr gem

data = Parsr.literal_eval(str)

You could also write your own grammar, but... why? :)