Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
296 views
in Technique[技术] by (71.8m points)

node.js - Stub module function called from the same module

I can't figure out a way to stub a function called from within the same module this function is defined (the stub does not seem to work). Here's an example:

myModule.js:

'use strict'

function foo () {
  return 'foo'
}

exports.foo = foo

function bar () {
  return foo()
}

exports.bar = bar

myModule.test.js:

'use strict'

const chai = require('chai')
const sinon = require('sinon')

chai.should()

const myModule = require('./myModule')

describe('myModule', () => {
  describe('bar', () => {
    it('should return foo', () => {
      myModule.bar().should.equal('foo') // succeeds
    })

    describe('when stubbed', () => {
      before(() => {
        sinon.stub(myModule, 'foo').returns('foo2') // this stub seems ignored
      })

      it('should return foo2', () => {
        myModule.bar().should.equal('foo2') // fails
      })
    })
  })
})

This reminds me of Java static functions which are not stubbable (almost).

Any idea how to achieve what I'm trying to do? I know that extracting foo in a different module will work, but that's not what I'm trying to do here. I'm also aware that invoking foo in the bar method with the keyword this will also work, I'm puzzled toward the use of ?this?in this context (since I'm not using OOP).

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I just tested this. And it works like charm.

'use strict'

function foo () {
  return 'foo';
}

exports.foo = foo;

function bar () {
  return exports.foo(); // <--- notice
}

exports.bar = bar;

Explanation

when you do sinon.stub(myModule, 'foo').returns('foo2') then sinon stubs the exported object's foo not the actually foo function from inside your myModule.js ... as you must know, foo is in accessible from outside the module. So when you set exports.foo, the exported object exports.foo stores the ref of foo. and when you call sinon.stub(myModule, 'foo').returns('foo2'), sinon will stub exports.foo and not the actual foo

Hope this makes sense!


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...