Question

具有以下balanceRepository.js文件。

module.exports = function(){
this.getBalance = (accountId)=>{
    throw 'NotImplementedException';
};};

还有服务transactionService.js文件。

module.exports = function(balanceRepo){
this.isTransactionAllowed = (amount, accountId)=>{
    const balance = balanceRepo.getBalance(accountId);
    return amount <= balance;
};};

我想进行单元测试，当我们调用transactionService.isTransactionAllowed时，它将在内部调用balanceRepository.getBalance。我期待以下代码能正常工作。

let BalanceRepository = require('../balanceRepository');
let TransactionService = require('../transactionService');

let should = require('should');
let sinon = require('sinon');

describe('transaction service', ()=>{
   let fakeBalanceRepo = sinon.spy(BalanceRepository);
   let transactionSrv = new TransactionService(fakeBalanceRepo);

   transactionSrv.isTransactionAllowed(100, 987654321);

   it('should call getBalance', ()=>{
      fakeBalanceRepo.getBalance.should.be.called();
});});

我创建了类BalanceRepository的间谍。我将那个间谍对象注入到我的TransactionService的实际实现中。然后，我断言在间谍程序上调用了getBalance方法。这对我来说很有意义，但是没有用！！

如何注入伪造品并断言其方法之一被称为？

Answer 1

sinon.spy(BalanceRepository)并不意味着会自动侦听类方法（不会）。

TransactionService期望BalanceRepository实例，而fakeBalanceRepo是构造函数。

如果目的是监视类方法而不改变其行为，则为：

   let balanceRepo = new BalanceRepository;
   sinon.spy(balanceRepo, 'isTransactionAllowed')
   let transactionSrv = new TransactionService(balanceRepo);
   transactionSrv.isTransactionAllowed(100, 987654321);
   balanceRepo.getBalance.should.be.called();

Answer 2

这是发情期所评论的。

let balanceRepo = new BalanceRepository();
let fakeGetBalance = sinon.stub(balanceRepo, 'getBalance');

let transactionSrv = new TransactionService(balanceRepo);
transactionSrv.isTransactionAllowed(100, 987654321);

it('should call getBalance', ()=>{
    fakeGetBalance.should.be.called();
});

用Sinon.js模拟要注入的依赖项

2 个答案: