javascript - sinon 监视功能不起作用

Question

我正在尝试为这个简单的中间件功能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
  })
})

但这确实..

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
  })
})

为什么监视功能不能正常工作？

Answer 1

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

因此，您的第一个示例等于:

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

您的第二个示例等于:

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，sinon 中“监视”和“ stub ”的关键是您必须在测试中使用监视/ stub 函数。您的原始代码仅使用原始的非 spy 功能。