如何干掉这个RSpec控制器规格？

Question

我怎么能干这个规格？

describe Api::TasksController, type: :controller do
  it 'allows the creator of a task to destroy it' do
    set_request_auth_header @user
    delete :destroy, id: @task.id
    expect(response).to be_success
    expect(Task.count).to eq 0
  end

  it 'does not allow the assignee of a task to destroy it' do
    set_request_auth_header @assignee
    delete :destroy, id: @task.id
    expect(response).to be_forbidden
    expect(Task.count).to eq 1
  end

  it 'does not allow anyone unrelated to a task to destroy it' do
    set_request_auth_header @spy
    delete :destroy, id: @task.id
    expect(response).to be_forbidden
    expect(Task.count).to eq 1
  end
end

Answer 1

你没有。 瞧，欢迎你。

Answer 2

你可以提取一些方法：

describe Api::TasksController, type: :controller do
  it 'allows the creator of a task to destroy it' do
    expect_delete_to_succeed @user
  end

  it 'does not allow the assignee of a task to destroy it' do
    expect_delete_to_be_forbidden @assignee
  end

  it 'does not allow anyone unrelated to a task to destroy it' do
    expect_delete_to_be_forbidden @spy
  end

  def expect_delete_to_succeed(requester)
    delete_task requester
    expect(response).to be_success
    expect(Task.count).to eq 0
  end

  def expect_delete_to_be_forbidden(requester)
    delete_task requester
    expect(response).to be_forbidden
    expect(Task.count).to eq 1
  end

  def delete_task(requester)
    set_request_auth_header requester
    delete :destroy, id: @task.id
  end

end

附注：

• 使用实例（@）变量不是RSpec中的当前标准实践。 考虑let “变量” 。
• Task.exists? @task.id Task.exists? @task.id似乎比检查Task.count更清楚地检查删除。