处理递归与异步函数和承诺的最佳方式?

Question

下面是伪代码来说明我的问题。当所有任务完成时，父函数最终必须返回一个promise(为了清晰起见，我省略了其他任务)。父函数调用子函数，一些子函数必须递归地执行它们的任务，因此，为了清晰起见，我将它们分离为辅助函数。如果有更干净的方法，我愿意学习。

在这个例子中如何最好地处理递归?

// This function must ultimately return a Promise.
async function parentFunction(uId) {
try {
await childFunction(uId);
return Promise.resolve(uId);
} catch (error) {
console.log(error);
}
}
async function childFunction(uId) {
try {
const done = await workerFunction(uId);
if (done) {
return Promise.resolve(true);
} else {
// There are more files to delete; best way to handle recursion?
}
} catch (error) {
console.log(error);
}
}
async function workerFunction(uId) {
try {
// Query the database, limit to 100 files.
const query = await db.queryFiles().limit(100);
if (query.size == 0) {
// Nothing to delete, we're done!
return Promise.resolve(true);
}
// Perform an atomic (all-or-none) batch delete that can only take 100 files at most.
await db.batchDelete(query);

// Batch delete successfull!
if (query.size < 100) {
// The query was less than 100 files so there can be no more files to delete.
return Promise.resolve(true);
} else {
// There may possibly be more files to delete.
// Return a promise or handle recursion here?
return Promise.resolve(false);
}
} catch (error) {
console.log(error);
}
}

Answer 1

只要递归就好了!

async function deleteFiles() {
const query = await db.queryFiles().limit(100)
if (query.size > 0) {
await db.batchDelete(query)
}
if (query.size === 100) {
return deleteFiles()
}
return true;
}