NodeJS Jest mock setTimeout -获取超时错误

km0tfn4u 于 2023-03-29 发布在 Node.js

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我正在写一个单元测试，它测试一个在继续之前等待超时的函数，我试过使用

jest.useFakeTimers();
jest.advanceTimersByTime(20000);

但我一直收到错误信息
: Timeout - Async callback was not invoked within the 20000 ms timeout specified by jest.setTimeout.Timeout - Async callback was not invoked within the 5000 ms timeout specified by jest.setTimeout.Error:

模块文件：

async function main()

{

   const promisesArray: Promise<void>[] = [];

   // lots of logic and async calls //

   for (const userRecord of records) {

      promisesArray.push(verifyUserRecord(userRecord.userId));

   }

   await Promise.all(promisesArray);
}

function waitTimeout(ms: number) {
     return new Promise((resolve) => setTimeout(() => resolve(true), ms));
}

 async function verifyUserRecord(userId) {
   
    await waitTimeout(CHECK_EXECUTION_STATUS_TIMEOUT);

    await <some other method call>
   /////////////// more logic /////////////
}

测试文件

describe('main test', () => {
beforeAll(() => {
  // calls to jest spy on //
});
beforeEach(() => {
  jest.useFakeTimers();
  // various mocks //

});
afterEach(() => {
  jest.resetAllMocks();
});
afterAll(() => {
  jest.useRealTimers();
  jest.restoreAllMocks();
});
it('Should successfully run', async () => {
    const p = main();
    jest.advanceTimersByTime(60000);
    jest.runAllTimers(); // <- explicitly tell jest to run all setTimeout, setInterval
    jest.runAllTicks(); // <- explicitly tell jest to run all Promise callback

    await p;

    // validations // 
  });
});

但是当我从测试文件调用main方法时，我得到了上面的错误。当调试时，我看到它到达waitTimeout，然后等待或移动到测试模块中的下一个it。它没有模拟超时，并继续模块文件逻辑。
如何正确测试文件并模拟超时？
先谢了。

node.js

来源：https://stackoverflow.com/questions/67959278/jest-mock-settimeout-getting-timeout-error

3条答案

按热度按时间

ozxc1zmp1#

模拟计时器时使用了错误的API
jest.setTimeout(20000);用于设置jest等待测试完成的最大超时。如果测试时间过长，jest将抛出错误
回到您的问题，因为您使用jest.useFakeTimers();，所以必须显式地告诉jest用尽所有宏任务（setTimeout）和微任务（Promise回调）。
我对你的代码做了一点修改，这样expect就可以工作了

const CHECK_EXECUTION_STATUS_TIMEOUT = 2000;
const records = [{ userId: 1 }, { userId: 2 }];

async function main() {
  const promisesArray: Promise<number>[] = [];

  for (const userRecord of records) {
    promisesArray.push(verifyUserRecord(userRecord.userId));
  }

  return await Promise.all(promisesArray);
}

function waitTimeout(ms: number) {
  return new Promise((resolve) => setTimeout(() => resolve(true), ms));
}

async function verifyUserRecord(userId: number) {
  await waitTimeout(CHECK_EXECUTION_STATUS_TIMEOUT);

  // await <some other method call>
  /////////////// more logic /////////////
  return userId;
}

describe("main", () => {
  beforeEach(() => {
    jest.useFakeTimers(); // <- use fake timer
  });

  afterEach(() => {
    jest.useRealTimers();
  });

  it("should work", async () => {
    const p = main();
    jest.runAllTimers(); // <- explicitly tell jest to run all setTimeout, setInterval
    jest.runAllTicks(); // <- explicitly tell jest to run all Promise callback

    expect(await p).toEqual([1, 2]);
  });
});

输出：

赞(0）回复(0）举报 2023-03-29

lhcgjxsq2#

您可以使用runOnlyPendingTimers，这将刷新所有挂起的计时器。
用途

jest.useFakeTimers();

// call your function here

jest.runOnlyPendingTimers();

赞(0）回复(0）举报 2023-03-29

8yparm6h3#

另一种模拟setTimeout的方法：

jest.spyOn(global, 'setTimeout').mockImplementation((callback) => {
if (typeof callback === 'function') {
    callback();
}
return { hasRef: () => false } as NodeJS.Timeout;
});

赞(0）回复(0）举报 2023-03-29

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NodeJS Jest mock setTimeout -获取超时错误

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