sinon 间谍功能不起作用

Question

我正在尝试为这个简单的中间件函数编写一个独立的测试

function onlyInternal (req, res, next) {
  if (!ReqHelpers.isInternal(req)) {
    return res.status(HttpStatus.FORBIDDEN).send()
  }
  next()
}

// Expose the middleware functions
module.exports = {
  onlyInternal
}

这不起作用

describe('success', () => {

  let req = {
    get: () => {return 'x-ciitizen-token'}
  }
  let res = {
    status: () => {
      return {
        send: () => {}
      }
    }
  }
  function next() {}
  let spy

  before(() => {
    spy = sinon.spy(next)
  })

  after(() => {
    sinon.restore()
  })

  it('should call next', () => {
    const result = middleware.onlyInternal(req, res, next)
    expect(spy.called).to.be.true <-- SPY.CALLED IS ALWAYS FALSE EVEN IF I LOG IN THE NEXT FUNCTION SO I KNOW IT'S GETTING CALLED
  })
})

但确实如此..

describe('success', () => {

  let req = {
    get: () => {return 'x-ciitizen-token'}
  }
  let res = {
    status: () => {
      return {
        send: () => {}
      }
    }
  }
  let next = {
    next: () => {}
  }
  let spy

  before(() => {
    spy = sinon.spy(next, 'next')
  })

  after(() => {
    sinon.restore()
  })

  it('should call next', () => {
    const result = middleware.onlyInternal(req, res, next.next)
    expect(spy.called).to.be.true
  })
})

为什么不能只监视函数工作？

Answer 1

Sinon 无法更改现有函数的内容，因此它创建的所有间谍都只是现有函数的包装器，用于计算调用次数、记忆参数等。

所以，你的第一个例子是这样的：

function next() {}
let spy = sinon.spy(next);
next(); // assuming that middleware is just calling next
// spy is not used!

你的第二个例子，等于：

let next = { next: () => {} }
next.next = sinon.spy(next.next); // sinon.spy(obj, 'name') just replaces obj.name with spy on it
next.next(); // you actually call spy which in in turn calls original next.next
//spy is called. YaY

因此，在 sinon 中“spying”和“stubbing”的关键在于您必须在测试中使用 spied/stubbed 函数。您的原始代码只是使用了原始的、非间谍功能。