Question

在Rails应用程序中，我有一个调用服务对象的后台作业

class MyJob < ActiveJob::Base
  def perform( obj_id )
    return unless object = Object.find(obj_id)
    MyServiceObject.new(object).call
  end
end

我可以测试该作业如下调用服务对象：

describe MyJob, type: :job do
  let(:object) { create :object }
  it 'calls MyServiceObject' do
    expect_any_instance_of(MyServiceObject).to receive(:call)
    MyJob.new.perform(object)
  end
end

但是如何测试作业是否使用正确的参数初始化服务对象？

describe MyJob, type: :job do
  let(:object) { create :object }
  it 'initializes MyServiceObject with object' do
    expect( MyServiceObject.new(object) ).to receive(:call)
    MyJob.new.perform(object)
  end
end

我希望实现上述内容，但这会失败expects 1 but received 0。

正确初始化测试类的正确方法是什么？

Answer 1

显然我对Adilbiy的答案的解决方案被拒绝了，所以这是我的最新答案：

describe MyJob, type: :job do
  let(:object) { create :object }
  it 'initializes MyServiceObject with object' do
    so = MyServiceObject.new(object)
    expect(MyServiceObject).to receive(:new).with(object).and_return(so)
    MyJob.new.perform(object)
  end
end

通过使用MyServiceObject.new模拟expect(MyServiceObject).to receive(:new).with(object)，您将覆盖原始实现。但是，要使MyJob#perform生效，我们需要MyServiceObject.new来返回一个对象 - 您可以使用.and_return(...)执行此操作。

希望这有帮助！

Answer 2

我可能会建议进行以下测试：

describe MyJob, type: :job do
  let(:object) { create :object }
  it 'initializes MyServiceObject with object' do
    expect( MyServiceObject).to receive(:new).with(object)
    MyJob.new.perform(object)
  end
end

如何使用正确的参数初始化该服务对象？

2 个答案: