tto*_*ous 5 javascript node.js promise typescript
我想要一些 JavaScript 代码将 3 个东西作为参数:
我最终做的是使用for循环。我不想使用递归函数:这样,即使有 50 次尝试,调用堆栈也不会长 50 行。
这是代码的打字稿版本:
/**
* @async
* @function tryNTimes<T> Tries to resolve a {@link Promise<T>} N times, with a delay between each attempt.
* @param {Object} options Options for the attempts.
* @param {() => Promise<T>} options.toTry The {@link Promise<T>} to try to resolve.
* @param {number} [options.times=5] The maximum number of attempts (must be greater than 0).
* @param {number} [options.interval=1] The interval of time between each attempt in seconds.
* @returns {Promise<T>} The resolution of the {@link Promise<T>}.
*/
export async function tryNTimes<T>(
{
toTry,
times = 5,
interval = 1,
}:
{
toTry: () => Promise<T>,
times?: number,
interval?: number,
}
): Promise<T> {
if (times < 1) throw new Error(`Bad argument: 'times' must be greater than 0, but ${times} was received.`);
let attemptCount: number;
for (attemptCount = 1; attemptCount <= times; attemptCount++) {
let error: boolean = false;
const result = await toTry().catch((reason) => {
error = true;
return reason;
});
if (error) {
if (attemptCount < times) await delay(interval);
else return Promise.reject(result);
}
else return result;
}
}
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上面使用的函数delay是一个约定的超时:
/**
* @function delay Delays the execution of an action.
* @param {number} time The time to wait in seconds.
* @returns {Promise<void>}
*/
export function delay(time: number): Promise<void> {
return new Promise<void>((resolve) => setTimeout(resolve, time * 1000));
}
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澄清一下:上面的代码有效,我只是想知道这是否是一种“好的”方法,如果不是,我该如何改进它。
有什么建议吗?在此先感谢您的帮助。
我不想使用递归函数:这样,即使有 50 次尝试,调用堆栈也不会长 50 行。
这不是一个好的借口。调用堆栈不会因异步调用而溢出,并且当递归解决方案比迭代解决方案更直观时,您可能应该采用它。
我最终做的是使用
for循环。这是一种“好”的做法吗?如果不是,我该如何改进?
循环for很好。虽然它的开始有点奇怪1,但基于 0 的循环更惯用。
然而,不好的是你奇怪的错误处理。该布尔error标志不应在您的代码中占有一席之地。使用.catch()很好,但try/ 也catch同样有效,应该是首选。
export async function tryNTimes<T>({ toTry, times = 5, interval = 1}) {
if (times < 1) throw new Error(`Bad argument: 'times' must be greater than 0, but ${times} was received.`);
let attemptCount = 0
while (true) {
try {
const result = await toTry();
return result;
} catch(error) {
if (++attemptCount >= times) throw error;
}
await delay(interval)
}
}
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