尝试使用httparty（Ruby）从Twitter获取朋友列表

Question

我正在尝试从Twitter获取特定用户朋友的列表。
这是我的代码 -

require 'rubygems'
require 'httparty'

class TwitterData
  include HTTParty
  base_uri 'http://api.twitter.com/1/'
  default_params :output => 'json'
  format :json

  def self.get_username_data(username)
    get('statuses/friends.json' , :query => { :screen_name => username })
  end
end

puts "Please your twitter username - "
twitter_username = gets
puts TwitterData.get_username_data(twitter_username).inspect

这是我遇到的错误 -

Please your twitter username -
twitter
C:/Ruby192/lib/ruby/gems/1.9.1/gems/crack-0.1.8/lib/crack/json.rb:14:in `rescue in parse': Invalid JSON string (Crack::ParseError)
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/crack-0.1.8/lib/crack/json.rb:12:in `parse'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty/parser.rb:116:in `json'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty/parser.rb:136:in `parse_supported_format'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty/parser.rb:103:in `parse'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty/parser.rb:66:in `call'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty/request.rb:180:in `parse_response'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty/request.rb:164:in `handle_response'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty/request.rb:57:in `perform'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty.rb:280:in `perform_request'
        from C:/Ruby192/lib/ruby/gems/1.9.1/gems/httparty-0.6.1/lib/httparty.rb:232:in `get'
        from twitter_friends_2.rb:11:in `get_username_data'
        from twitter_friends_2.rb:17:in `<main>'

Answer 1

用这个替换您的方法

def self.get_username_data(username)
  get("/statuses/friends.json?screen_name=#{username}")
end

或者

def self.get_username_data(username)
  get('/statuses/friends.json' , :query => { :screen_name => username.strip })
end

您需要剥离用户名，因为当用户输入Twitter用户名时，从命令行中输入键A“ n”会附加到用户名上，并且相同的用户名是作为引起问题的参数发送。以上任何给定的代码段都应起作用。