我在用jest为下面的代码片段编写单元测试时遇到了困难：

async addCronJob(props: IAddAndUpdateCronJobDetails) {
   const {name, individualSchedule} = props;
   const parsedCronTime = convertDateAndTimeToCron(
   individualSchedule.timeOfRun,
   individualSchedule.dateOfrun
   )

  const {jobType, dateOfRun, id, timeOfRun} = individualSchedule;

  const newJob = new CronJob(
   parsedCronTime,
   async () => {
   return this.sqsService.getSqsApproval({
   //some properties
    }).then(() => {
    //some logic
    })
   },
   null,
   false,
   'Asia/Singapore'
  )

 this.schedulerRegistry.addCronJob(name, newJob)
 newJob.start()
}

下面是我的单元测试：

//at the top
jest.mock('cron', () => {
const mScheduleJob = {start: jest.fn(), stop: jest.fn()};
const mCronJob = jest.fn(() => mScheduleJob);
return {CronJob: mCronJob}
})

***************

describe('addCronJob', () => {
 it('should add a new cron job', async (done) => {
  const testFn = jest.fn();
  const parsedCronTime = convertDateAndTimeToCron(
   mockSchedule.timeOfRun,
   mockSchedule.dateOfrun
   )
  const testCronJob = new CronJob(
  parsedCronTime,
  testFn,
  null,
  false,
  'Asia/Singapore'
  );
 
 return dynamicCronService.addCron({//properties}).then(() => {
   expect(CronJob).toHaveBeenCalledWith(//properties);
   expect(testCronJob.start).toBeCalledTimes(1);
   done()
 })
 })

})

上述测试通过，没有错误。但是，无法在cron作业本身中测试此异步代码块：

async () => {
   return this.sqsService.getSqsApproval({
   //some properties
    }).then(() => {
    //some logic
    })
 }

有人知道如何测试上面的代码块吗？
谢谢!