javascript - 如何模拟在 AngularJS Jasmine 单元测试中返回 promise 的服务？

Question

我有 myService 使用 myOtherService，它进行远程调用，返回 promise :

angular.module('app.myService', ['app.myOtherService'])
  .factory('myService', [
    myOtherService,
    function(myOtherService) {
      function makeRemoteCall() {
        return myOtherService.makeRemoteCallReturningPromise();
      }

      return {
        makeRemoteCall: makeRemoteCall
      };      
    }
  ])

要对 myService 进行单元测试，我需要模拟 myOtherService，使其 makeRemoteCallReturningPromise 方法返回一个 promise 。我就是这样做的:

describe('Testing remote call returning promise', function() {
  var myService;
  var myOtherServiceMock = {};

  beforeEach(module('app.myService'));

  // I have to inject mock when calling module(),
  // and module() should come before any inject()
  beforeEach(module(function ($provide) {
    $provide.value('myOtherService', myOtherServiceMock);
  }));

  // However, in order to properly construct my mock
  // I need $q, which can give me a promise
  beforeEach(inject(function(_myService_, $q){
    myService = _myService_;
    myOtherServiceMock = {
      makeRemoteCallReturningPromise: function() {
        var deferred = $q.defer();

        deferred.resolve('Remote call result');

        return deferred.promise;
      }    
    };
  }

  // Here the value of myOtherServiceMock is not
  // updated, and it is still {}
  it('can do remote call', inject(function() {
    myService.makeRemoteCall() // Error: makeRemoteCall() is not defined on {}
      .then(function() {
        console.log('Success');
      });    
  }));  

从上面可以看出，我的 mock 的定义取决于 $q，我必须使用 inject() 加载它。此外，注入(inject)模拟应该发生在 module() 中，它应该在 inject() 之前发生。但是，一旦我更改了模拟的值，它就不会更新。

这样做的正确方法是什么？

Answer 1

我不知道为什么你这样做的方式不起作用，但我通常用 spyOn 函数来做。像这样的:

describe('Testing remote call returning promise', function() {
  var myService;

  beforeEach(module('app.myService'));

  beforeEach(inject( function(_myService_, myOtherService, $q){
    myService = _myService_;
    spyOn(myOtherService, "makeRemoteCallReturningPromise").and.callFake(function() {
        var deferred = $q.defer();
        deferred.resolve('Remote call result');
        return deferred.promise;
    });
  }

  it('can do remote call', inject(function() {
    myService.makeRemoteCall()
      .then(function() {
        console.log('Success');
      });    
  }));

还请记住，您需要调用 $digest 才能调用 then 函数。请参阅 $q documentation 的 测试 部分.

-----编辑------

仔细查看您的操作后，我认为您的代码中存在问题。在 beforeEach 中，您将 myOtherServiceMock 设置为一个全新的对象。 $provide 永远不会看到这个引用。您只需要更新现有的引用:

beforeEach(inject( function(_myService_, $q){
    myService = _myService_;
    myOtherServiceMock.makeRemoteCallReturningPromise = function() {
        var deferred = $q.defer();
        deferred.resolve('Remote call result');
        return deferred.promise;   
    };
  }