JEST:如何对在类的构造函数中调用的方法进行存根

问题描述：

示例如下:

export class ABC {
  constructor() {
    this.method1();
  }

  method1() {
    console.log();
  }
}

假定method1中有一些对外部方法的调用，这些调用正在停止前进的代码.我不想进入method1.

Assume that there are some calls to external methods in method1 which are stopping the code to go forward. I don't want to go inside the method1.

现在的问题是当我这样做时:

Now the problem is when I do this:

describe('test cases!', () => {
  let abc: ABC;
  beforeEach(() => {
    spyOn(abc, 'method1').and.stub();
    abc = new ABC();
    jest.resetAllMocks();
  });
});

抛出错误.

在类初始化之后，我不能放置spyOn. 有想法的人吗?

After the class initialisation I can't put the spyOn. Any idea guys?

感谢帮助.

答

要在@ brian-lives-outdoors答案的基础上，我使用以下内容:

To build upon @brian-lives-outdoors answer, I use the following:

describe('test cases!', () => {
  beforeEach(() => {
    ABC.prototype.abc = jest.fn((data) => { result: 'some mock here', data)
    abc = new ABC();
  });
  afterEach(() => {
    ABC.prototype.abc.mockRestore()
  })
});

JEST:如何对在类的构造函数中调用的方法进行存根

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