我需要存根具有特定属性或属性集的模型的所有实例.例如,使用ActiveRecord:
let(:model1) { Model.create!(uid: 1) }
let(:model2) { Model.create!(uid: 2) }
before do
allow(model1).to receive(:foo).and_return :bar
allow(model2).to receive(:foo).and_return :baz
end
it do
expect(model1.foo).to eq :bar # => passes
expect(model2.foo).to eq :baz # => passes
################################################
#### Here is the issue I'm trying to solve: ####
################################################
new_instance_of_model1 = Model.find_by(uid: 1)
new_instance_of_model2 = Model.find_by(uid: 2)
expect(new_instance_of_model1.foo).to eq :bar # => fails
expect(new_instance_of_model2.foo).to eq :baz # => fails
end
有没有办法将所有具有uid的Model实例存根:1?
我正在寻找类似的东西:
allow_any_instance_of(Model).with_attributes(uid: 1).to receive(:foo).and_return(:bar)
allow_any_instance_of(Model).with_attributes(uid: 2).to receive(:foo).and_return(:baz)
注意:
我不能使用像:
allow(Model).to receive(:find).with(1)and_return(model1)
allow(Model).to receive(:find).with(2)and_return(model2)
因为还有很多其他方法可以进入模型(协会,范围,Arel等)
最佳答案 这不是一个好主意,但是您可以通过将块参数传递给receive而不是方法链来完成此操作,如下所示:
allow_any_instance_of(Model).to receive(:foo) do |model|
case model.uid
when 1
:bar
when 2
:baz
end
end