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我想测试一个直接呈现一些 JSON 输出的控制器(通过使用“render :json => @entity_names”)。对于该任务,我在我的规范文件“response.should have_text('["enim", "enita"]')" 中尝试过。不幸的是,我总是得到那个错误:Failure/Error: response.should have_text('["enim", "enita"]') undefined method `has_text?' 为了 #

我是否错过了一些提供该方法的宝石?这是我的 Gemfile:

source 'http://rubygems.org'

gem 'rails', '>= 3.0.0'
gem 'mysql2'
gem 'mongrel'
gem 'devise'
gem 'will_paginate', :git => 'git://github.com/mislav/will_paginate.git', :branch =>    'rails3'
gem 'thinking-sphinx', :git     => 'git://github.com/freelancing-god/thinking-sphinx.git', :branch  => 'rails3', :require => 'thinking_sphinx'

group :test, :development do
  gem 'rspec-rails', '>= 2.0.0.beta.19'
  gem 'steak', :git => 'git://github.com/cavalle/steak.git'
  gem 'webrat'
  gem 'capybara'
  gem 'capybara-envjs'
  gem 'shoulda'
  gem 'launchy'
  gem 'autotest'
  gem 'autotest-rails'
  gem 'test_notifier'
  gem 'rails3-generators'
  gem 'factory_girl_rails'
  gem 'populator'
  gem 'faker'
  gem 'random_data'
  gem 'database_cleaner', :git => 'git://github.com/bmabey/database_cleaner.git'
  gem 'delorean'
end
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1 回答 1

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您可以将预期的输出构造为 JSON,然后获取响应正文(也是 JSON),对两者进行解码并进行比较。就像是:

it "should do something" do
  expected = { :some_key => "and some value" }.to_json 
  xhr :post, :create, { :foo => "bar" }
  ActiveSupport::JSON.decode(response.body).should == ActiveSupport::JSON.decode(expected)
end 
于 2010-11-21T23:56:26.220 回答