Sinon间谍在console.log上调用未注册

Question

我正在努力了解Sinon,并希望窥探console.log.代码很简单:

function logToConsole() {
    console.log('Hello World');
}

exports.logToConsole = logToConsole;

但是,如果我想测试它,它不起作用,因为调用console.log没有在被测系统内注册:

var chai = require('chai'),
    expect = chai.expect,
    sinonChai = require('sinon-chai'),
    sinon = require('sinon'),
    sut = require('../src/logToConsole');

chai.use(sinonChai);

describe('logToConsole', function() {
    it('should spy on console.log', function() {
        sinon.spy(console, 'log');

        sut.logToConsole();

        expect(console.log).to.have.been.called;
    });
});

但是,如果我console.log在测试本身内部执行,则会捕获并传递:

it('should spy on console.log', function() {
    sinon.spy(console, 'log');

    sut.logToConsole();
    console.log('Test');

    expect(console.log).to.have.been.called;
});

有趣的是,它似乎根本无法监视内部函数调用.这不是间谍图书馆的目的吗？

例如

function a() {};

function b() {
    a();
}

Answer 1

看起来你实际上并没有使用sinon-chai,你发布的代码缺少这一行:

chai.use(sinonChai);

编辑:这是我测试的代码:

// test.js
var chai = require('chai'),
    expect = chai.expect,
    sinonChai = require('sinon-chai'),
    sinon = require('sinon'),
    sut = require('./log');

chai.use(sinonChai);

describe('logging', function() {

  beforeEach(function() {
    sinon.spy(console, 'log');
  });

  afterEach(function() {
    console.log.restore();
  });

  describe('logToConsole', function() {
    it('should log to console', function() {
      sut.logToConsole();
      expect(console.log).to.be.called;
    });
  });

  describe('logToConsole2', function() {
    it('should not log to console', function() {
      sut.logToConsole2();
      expect(console.log).to.not.be.called;
    });
  });
});

// log.js
module.exports.logToConsole = function() {
  console.log('Hello World');
};

module.exports.logToConsole2 = function() {
};