我该如何模拟在AngularJS Jasmine单元测试中返回诺言的服务？

Question

我有myService使用myOtherService，它可以进行远程调用，并返回promise：

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

      return {
        makeRemoteCall: makeRemoteCall
      };      
    }
  ])

要对myService我进行单元测试，需要模拟myOtherService，以便其makeRemoteCallReturningPromise方法返回promise。这是我的方法：

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');
      });    
  }));

从上面可以看到，我的模拟的定义取决于$q，我必须使用进行加载inject()。此外，应该在中进行注入模拟module()，这应该在之前进行inject()。但是，更改模拟后，其值不会更新。

正确的方法是什么？

Answer 1

我不确定为什么您的方法行不通，但是我通常使用该spyOn函数来完成。像这样：

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');
      });    
  }));

还要记住，您将需要$digest调用要调用的then函数。请参阅$
q文档
的“ 测试”
部分。

- - - 编辑 - - -

在仔细研究您的操作之后，我认为我在您的代码中看到了问题。在中beforeEach，您将设置myOtherServiceMock为一个新对象。将$provide永远不会看到这个参考。您只需要更新现有参考：

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