How do I parse JSON with Ruby on Rails? [duplicate]

These answers are a bit dated. Therefore I give you:

hash = JSON.parse string

Rails should automagically load the json module for you, so you don't need to add require 'json'.

Parsing JSON in Rails is quite straightforward:

parsed_json = ActiveSupport::JSON.decode(your_json_string)

Let's suppose, the object you want to associate the shortUrl with is a Site object, which has two attributes - short_url and long_url. Than, to get the shortUrl and associate it with the appropriate Site object, you can do something like:

parsed_json["results"].each do |longUrl, convertedUrl|
  site = Site.find_by_long_url(longUrl)
  site.short_url = convertedUrl["shortUrl"]
  site.save
end

This answer is quite old. pguardiario's got it.

One site to check out is JSON implementation for Ruby. This site offers a gem you can install for a much faster C extension variant.

With the benchmarks given their documentation page they claim that it is 21.500x faster than ActiveSupport::JSON.decode

The code would be the same as Milan Novota's answer with this gem, but the parsing would just be:

parsed_json = JSON(your_json_string)

Here is an update for 2013.

Ruby

Ruby 1.9 has a default JSON gem with C extensions. You can use it with

require 'json'
JSON.parse ''{ "x": "y" }'
# => {"x"=>"y"}

The parse! variant can be used for safe sources. There are also other gems, which may be faster than the default implementation. Please refer to multi_json for the list.

Rails

Modern versions of Rails use multi_json, a gem that automatically uses the fastest JSON gem available. Thus, the recommended way is to use

object = ActiveSupport::JSON.decode json_string

Please refer to ActiveSupport::JSON for more information. In particular, the important line in the method source is

data = MultiJson.load(json, options)

Then in your Gemfile, include the gems you want to use. For example,

group :production do
  gem 'oj'
end