Typescript 类中的类和函数的Jest模拟类实例

我有一个正在测试的类，我们称它为ToTest.ts，它实例化了另一个类的实例unrelated.ts，并调用了一个名为doSomething的方法

// ToTest.ts
const irrelevant = new Irrelevant();
export default class ToTest {

  // ... some implementation, constructor, etc.

  public static callIrrelevant() {
    return irrelevant.doSomething();
  }
}

//Irrelevant.ts
export default class Irrelevant {

  // ... some implementation, constructor, etc.

  public doSomething() {
    console.log("doing something");
  }
}

如何在我的jest测试中模拟无关的.doSomething（）

我试过：

import Irrelevant from '../some/path/irrelevant';
test('irrelevant.doSomething is mocked' => {
  Irrelevant.prototype.doSomething = () => jest.fn().mockImplementation(() => {
    console.log(`function got called`)
  });
  const toTest = new ToTest();
  toTest.callIrrelevant();
});
test('irrelevant.doSomething is mocked -- second try' => {
  jest.mock('../some/path/irrelevant')
  const mockDoSomething = jest.spyOn(Irrelevant, "doSomething"); // Gives error: Argument of type '"doSomething"' is not assignable to parameter of type 'never'.
  findMock.mockImplementation(() => {
    console.log(`function got called`)
  });
  const toTest = new ToTest();
  toTest.callIrrelevant();
 });
});

---

这两个测试都不会触发模拟实现。我怎样才能正确地模拟这个问题？

< P>你必须把McCK实现看作是<代码>的构造函数无关的< /代码>。它将返回什么将是你的嘲笑。所以你可以这样做：

import Irrelevant from '../some/path/irrelevant';

jest.mock('../some/path/irrelevant')

const spyDoSomething = jest.fn();

(Irrelevant as any).mockImplementation(() => {
  return { doSomething: spyDoSomething };
})

---

属性“mockImplementation”不存在于类型“typeof Unrelated”上。

我更新了答案。我只是对any进行了不相关的强制转换，因为它无论如何都会被模拟。您可以使用

ts jest

中的

mocked（）

方法：参见

import Irrelevant from '../some/path/irrelevant';
test('irrelevant.doSomething is mocked' => {
  Irrelevant.prototype.doSomething = () => jest.fn(() => {
    console.log(`function got called`)
  });
  const toTest = new ToTest();
  toTest.callIrrelevant();
});