如何用JEST测试promise递归

Question

null

null

export function retry <T> (fn: () => Promise <T>, limit: number = 5, interval: number = 10): Promise <T> {
  return new Promise ((resolve, reject) => {
    fn ()
      .then (resolve)
      .catch ((error) => {
        setTimeout (async () => {
          // Reject if the upper limit number of retries is exceeded
          if (limit === 1) {
            reject (error);
            return;
          }
          // If it is less than the upper limit number of retries, execute callback processing recursively
          await retry (fn, limit-1, interval);
        }, interval);
      });
  });
}

对上述重试功能执行以下测试。

1. 总是传递一个承诺进行解析，重试函数在第一次执行时解析/li> 解析重试函数 i="">

我以为写这些开玩笑的时候会是这样的。

describe ('retry', () => {
  test ('resolve on the first call', async () => {
    const fn = jest.fn (). mockResolvedValue ('resolve!');
    await retry (fn);
    expect (fn.mock.calls.length) .toBe (1);
  });

  test ('resolve on the third call', async () => {
    const fn = jest.fn ()
               .mockRejectedValueOnce (new Error ('Async error'))
               .mockRejectedValueOnce (new Error ('Async error'))
               .mockResolvedValue ('OK');
    expect (fn.mock.calls.length) .toBe (3)
  });
});

因此，它失败，出现以下错误。

Timeout-Async callback was not invoked within the 5000 ms timeout specified by jest.setTimeout.Error:
    > 40 | test ('resolve on the third call', async () => {
         | ^
      41 | const fn = jest
      42 | .fn ()
      43 | .mockRejectedValueOnce (new Error ('Async error'))

我认为，这将是可以管理的设置在笑话关于这个错误。但是，从根本上说，我不知道如何在Jest中测试promise递归处理。

Answer 1

超时是因为 中的 处理程序在第二次尝试调用 时不调用 ；因此，第一个 返回一个承诺，该承诺从未被解析或拒绝。

将 替换为 可能会有帮助（并且 中的功能不需要异步）:

  .catch ((error) => {
    setTimeout (() => {
      // Reject if the upper limit number of retries is exceeded
      if (limit === 1) {
        reject (error);
        return;
      }
      // If it is less than the upper limit number of retries, execute callback processing recursively
      resolve(retry (fn, limit-1, interval));
    }, interval);

Answer 2

null

可以使用 增加JEST的

null

我的解决方案是:

test("resolve on the third call", async () => {
    jest.setTimeout(10000);
    const fn = jest.fn()
      .mockRejectedValueOnce(new Error("Async error"))
      .mockRejectedValueOnce(new Error("Async error"))
      .mockResolvedValue("OK");

    // test reject value
    await expect(fn()).rejects.toEqual(new Error("Async error"));
    await expect(fn()).rejects.toEqual(new Error("Async error"));

    // test resolve
    const result = await fn();
    expect(result).toEqual("OK");

    // call time
    expect(fn).toHaveBeenCalledTimes(3);
  });