使用异步函数和承诺处理递归的最佳方法？

Question

以下是说明我的问题的伪代码。当所有任务完成时，父函数最终必须 return 一个承诺（为清楚起见，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，为了清楚起见，我将它们分成辅助函数。如果有更简洁的方法我很想学习。

如何最好地处理此示例中的递归？

// 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

递归就行了！

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;
}

使用异步函数和承诺处理递归的最佳方法？

Best way to handle recursion with async functions and promises?

javascript

recursion