单元测试解析器时如何模拟ActivatedRouteSnapshot

Question

我想为我的解析器编写一个单元测试,需要ActivatedRouteSnapshot像下面那样接受它的构造函数:

export class MyResolver {
  constructor () {
    // ...
  }

  resolve (private route: ActivatedRouteSnapshot) {
    callFoo(route.params.val);
  }
};

但在我的单元测试中,用模拟数据提供激活路由快照的最佳方法是什么？当我尝试使用我需要的属性创建一个对象时,我收到一个错误,它无法转换为ActivatedRouteSnapshot:

it('should call foo', inject([MyResolver], async (myResolver: MyResolver) => {
  const mockRoute = {
    params: {
      val: '1234'
    };
  };

  sinon.spy(callFoo);
  myResolver.resolve(mockRoute); // This is the line that errors
  expect(callFoo.calledWith('1234')).to.be.true;
}));

错误:

Type '{ params: { val: string; }; }' cannot be converted to type 'ActivatedRouteSnapshot'.

如何提供模拟ActivatedRouteSnapshot传递给我的解析器？

Answer 1

您必须像这样提供活动路线：

import {TestBed} from '@angular/core/testing';
import {SomeResolver} from './some.resolver';
import {ActivatedRoute, convertToParamMap} from '@angular/router';
import {of} from 'rxjs';

describe('SomeResolver', () => {
  let someResolver: SomeResolver;
  let route: ActivatedRoute;

  TestBed.configureTestingModule({
    providers: [
      {
        provide: ActivatedRoute,
        useValue: {snapshot: {paramMap: convertToParamMap({id: 'one-id'})}}
      },
      SomeResolver
    ]
  });

  beforeEach(() => {
    heroResolver = TestBed.get(HeroResolver);
    route = TestBed.get(ActivatedRoute);
  });

  it('should resolve', (() => {
    someResolver.resolve(route.snapshot);
  }));
});

在此处查看更复杂的示例。

Answer 2

我不确定这是否是最漂亮的解决方案，但您可以像这样模拟路线：

let route = createSpyObj('Route', ['']);
route.params = {
  val: '1234'
}