小编典典

如何在AngularJS Jasmine单元测试中模拟返回承诺的服务?

javascript

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,我必须使用来加载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?


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2020-05-01

共1个答案

小编典典

我不确定为什么您的方法不起作用,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 - - -

在仔细查看您在做什么之后,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;   
    };
  }
2020-05-01