Question

我有一个递归方法，我需要测试方法调用的顺序，以及调用它们的次数。我遇到的问题是，这些方法调用备用。例如......

it 'call methods in order' do
  expect(@instance).to have_received(:foo).ordered
  expect(@instance).to have_received(:bar).ordered
  expect(@instance).to have_received(:foo).ordered
  expect(@instance).to have_received(:bar).ordered
  expect(@instance).to have_received(:foo).ordered
  expect(@instance).to have_received(:bar).ordered
end

因为

而失败

expected: 1 time with any arguments
received: 3 times with any arguments

被调用3次的{p> foo和bar是正确的，但我如何交替测试它们？

Answer 1

这样的事情应该有效：

it 'call methods in order' do
  @method_calls = []

  allow(@instance).to receive(:foo) do
    @method_calls << :foo
  end

  allow(@instance).to receive(:bar) do
    @method_calls << :bar
  end

  ...

  expect(@method_calls).to eq([:foo, :bar, :foo, :bar, :foo, :bar])     
end

这是我尝试过的最小化测试，它充分展示了这一点：

require "rspec"

class MyClass
  def bar
  end

  def foo
  end

  def foobar
    foo
    bar
    foo
    bar
    foo
    bar
  end
end

describe "Foo" do
  describe "#foobar" do
    let(:instance) { MyClass.new }

    it "alternately calls foo/bar" do
      @method_calls = []

      allow(instance).to receive(:foo) do
        @method_calls << :foo
      end

      allow(instance).to receive(:bar) do
        @method_calls << :bar
      end

      instance.foobar

      expect(@method_calls).to eq([:foo, :bar, :foo, :bar, :foo, :bar])
    end
  end
end

运行此规范会通过，但删除（比方说）foobar方法中的一个方法调用会导致此错误：

F

Failures:

  1) Foo #foobar alternately calls foo/bar
     Failure/Error: expect(@method_calls).to eq([:foo, :bar, :foo, :bar, :foo, :bar])

       expected: [:foo, :bar, :foo, :bar, :foo, :bar]
            got: [:bar, :foo, :bar, :foo, :bar]

       (compared using ==)

       Diff:
       @@ -1,2 +1,2 @@
       -[:foo, :bar, :foo, :bar, :foo, :bar]
       +[:bar, :foo, :bar, :foo, :bar]
     # ./rspec_test.rb:38:in `block (3 levels) in <top (required)>'

Finished in 0.01517 seconds (files took 0.07714 seconds to load)
1 example, 1 failure

Failed examples:

rspec ./rspec_test.rb:25 # Foo #foobar alternately calls foo/bar

希望这是有道理的。

Answer 2

你应该封装方法的递归部分，存根其他所有内容，然后通过黑盒方法测试该方法（即输入 - >输出）。

如何测试有序和x次的rspec模拟？

2 个答案: