开玩笑spyOn（）调用实际函数，而不是模拟

Question

我正在测试apiMiddleware，它调用了其辅助函数callApi。为了防止调用实际的callApi来发出API调用，我对函数进行了模拟。但是，它仍然会被调用。

apiMiddleware.js

import axios from 'axios';

export const CALL_API = 'Call API';

export const callApi = (...arg) => {
  return axios(...arg)
    .then( /*handle success*/ )
    .catch( /*handle error*/ );
};

export default store => next => action => {
  // determine whether to execute this middleware
  const callAPI = action[CALL_API];
  if (typeof callAPI === 'undefined') {
    return next(action)
  }

  return callAPI(...callAPI)
    .then( /*handle success*/ )
    .catch( /*handle error*/ );
}

apiMiddleware.spec.js

import * as apiMiddleware from './apiMiddleware';

const { CALL_API, default: middleware, callApi } = apiMiddleware;

describe('Api Middleware', () => {

  const store = {getState: jest.fn()};
  const next = jest.fn();
  let action;

  beforeEach(() => {
    // clear the result of the previous calls
    next.mockClear();
    // action that trigger apiMiddleware
    action = {
      [CALL_API]: {
        // list of properties that change from test to test 
      }
    };
  });

  it('calls mocked version of `callApi', () => {
    const callApi = jest.spyOn(apiMiddleware, 'callApi').mockReturnValue(Promise.resolve());

    // error point: middleware() calls the actual `callApi()` 
    middleware(store)(next)(action);

    // assertion
  });
});

请忽略操作的属性和callApi函数的参数。我认为他们不是我要提出的要点。

如果需要进一步说明，请告诉我。