开玩笑的typescript属性嘲笑在类型上不存在 [英] jest typescript property mock does not exist on type

问题描述

使用jest.fn()添加模拟时，通常可以访问.mock属性以访问诸如呼叫之类的详细信息，类似于以下内容:

When using jest.fn() to add a mock you can usually access the .mock property to access details such as calls, something similar to this:

test('not working', () => {
    const foo = new Foo();
    foo.addListener = jest.fn();
    foo.func(); // will call addListener with a callback
    const callback = foo.addListener.mock.calls[0][0];
    expect(callback()).toEqual(1); // test the callback
});

当使用打字稿而不是普通的javascript实现测试时，出现错误:

When implementing the test in typescript instead of plain javascript I get the error:

错误TS2339:类型'((callback:()=> number)=> void')不存在属性'mock'.

error TS2339: Property 'mock' does not exist on type '(callback: () => number) => void'.

我可以通过强制转换为any来消除错误，但是肯定有更好的方法:

I can get rid of the error by casting to any but surely there must be a better way:

const callback = (foo.addListener as any).mock.calls[0][0];

在这个简单的代码中，可以使用jest.fn(fn => { callback = fn; });重写模拟程序以存储参数，但是使用foo.addListener.mockClear()时会发生相同的错误，无法以相同的方式进行修改.

In this simple code the mock could be rewritten to store the argument using jest.fn(fn => { callback = fn; }); but the same error happens when using foo.addListener.mockClear() which cannot be reworked the same way.

那么如何摆脱错误，最好在不丢失类型安全的情况下呢?

So how can I get rid of the error, preferably without losing type-safety?

推荐答案

您可以将 jest.spyOn 与<一个href ="https://jestjs.io/docs/en/mock-function-api#mockfnmockimplementationfn" rel ="noreferrer"> mockImplementation 来模拟一个函数，同时在TypeScript中保留类型安全性:

class Foo {
  addListener = (callback: () => number) => { }
  func = () => {
    this.addListener(() => 1);
  }
}

test('working', () => {
  const foo = new Foo();
  const mockAddListener = jest.spyOn(foo, 'addListener'); // spy on foo.addListener
  mockAddListener.mockImplementation(() => { }); // replace the implementation if desired
  foo.func(); // will call addListener with a callback
  const callback = mockAddListener.mock.calls[0][0];
  expect(callback()).toEqual(1); // SUCCESS
});

这篇关于开玩笑的typescript属性嘲笑在类型上不存在的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！