fre*_*gel 2 ruby rspec ruby-on-rails
我试图通过将一个哈希放入一个let并只调用let来干我的代码,但它似乎不起作用.我收到这些错误:
syntax error, unexpected =>, expecting '}' (SyntaxError) "email" => user.email,
syntax error, unexpected =>, expecting :: or '[' or '.' ...l" => user.email, "password" => user.password }
Run Code Online (Sandbox Code Playgroud)
这是我的测试:
describe '#create' do
let(:user) { create(:user) }
let(:user_params) { "email" => user.email, "password" => user.password }
before(:each) { User.stub(:find_by_email).and_return(user) }
it "should send find message to user model" do
User.should_receive(:find_by_email)
post :create, locale: "es", user: { user_params }
end
it "should set user to @user" do
post :create, locale: "es", user: { "email" => user.email, "password" => user.password }
assigns(:user).should eq(user)
end
end
Run Code Online (Sandbox Code Playgroud)
这是我的代码:
def create
@user = User.find_by_email(params[:email])
if @user && @user.authenticate(params[:password])
session[:user_id] = @user.id
redirect_to root_url
else
render 'login'
end
end
Run Code Online (Sandbox Code Playgroud)
Dog*_*ert 15
你错过了一对牙套.
let(:user_params) { "email" => user.email, "password" => user.password }
Run Code Online (Sandbox Code Playgroud)
应该
let(:user_params) { {"email" => user.email, "password" => user.password} }
Run Code Online (Sandbox Code Playgroud)
我更喜欢更长的解释.当你写这样的东西:
let(:user_params) { "email" => user.email, "password" => user.password }
Run Code Online (Sandbox Code Playgroud)
let使用一个参数(:user_params)和一个块(在本例中{ "email" => user.email, "password" => user.password })调用方法.我会再说一遍:
let(:something) { some_method_call }
Run Code Online (Sandbox Code Playgroud)
和
let(:something) do
some_method_call
end
Run Code Online (Sandbox Code Playgroud)
是平等的.通过使用let,RSpec将变量设置为块的结果.所以,在块内,你需要返回一些东西.在这种情况下 - 哈希:
let(:user_params) do
{ "email" => user.email, "password" => user.password }
end
Run Code Online (Sandbox Code Playgroud)
而已!您可以在APIdock上找到有关let的更多信息
| 归档时间: |
|
| 查看次数: |
6136 次 |
| 最近记录: |