javascript - 如何 stub 对象的属性而不是方法？

Question

我有一个文件 config.js，其中包含以下代码:

module.exports:
{
   development: {
        switch: false
        }
}

我有另一个文件bus.js，其中包含以下代码:

var config=require('config.js');

getBusiness:function(req,callback){
         if(config.switch) {
                  // Do something
         }else{
                 // Do something else
         }
}

现在，我想对文件bus.js进行单元测试

require('mocha');

var chai = require('chai'),
      expect = chai.expect,
      proxyquire = require('proxyquire');

var bus = proxyquire('bus.js', {
                 'config':{
                        switch:true
                  }
});

describe('Unit Test', function() {

        it('should stub the config.switch', function(done) {
            bus.getBusiness(req, function(data) {
              // It should stub the config.switch with true not false and give code coverage for if-else statmt. 
            });
           done();
        });
});

任何建议或帮助...

Answer 1

在我看来，您可以在测试文件中执行此操作:

var chai   = require('chai'),
  expect   = chai.expect;
var config = require('./config');

describe('Unit Test', function() {

  it('should stub the config.switch', function(done) {
    config.development.switch = true;
    bus.getBusiness(req, function(data) {
      ...
      done();
    });
  });

});