angular - 如何在对解析器进行单元测试时模拟 ActivatedRouteSnapshot

Question

我想为我的解析器编写一个单元测试，它需要在其构造函数中采用 ActivatedRouteSnapshot，如下所示:

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

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

但在我的单元测试中，使用模拟数据提供激活路线快照的最佳方式是什么？当我尝试创建一个只包含我需要的属性的对象时，我收到一个错误，指出它无法转换为 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'.

我如何提供模拟 ActivatedRouteSnapshot 以传递给我的解析器？

Answer 1

您必须像这样提供事件路线:

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);
  }));
});

检查一个更复杂的例子 here .