從 JavaScript 中的循環返回一個值？

Question

我正在創建一個 function 來刪除與我的數據庫中所需的“級別”數量不匹配的索引，然后返回該數組。

問題是當我執行cleanWinners(winners, guild).then(res => console.log(res))時，我收到未定義的消息。 我假設代碼在返回數組之前沒有等待這個 function 結束？ 我該如何解決？

當我使用這段代碼時：

async function cleanWinners(winners, guild) {
    let returnedArray;

    for (const winner of winners) {
        XP.findOne({serverID: guild.id, userID: winner}, (err, xpTable) => {
            if (err) throw err;
            if (!xpTable) {
                const newXP = new XP({
                    totalXP: 0,
                    xp: 0,
                    level: 0,
                    date: 0,
                    serverID: guild.id,
                    userID: winner
                });
                newXP.save().catch(console.log);
                xpTable = newXP;
            }
            // If they have less than the set amount, get a new winner.
            if (xpTable.level < 1) {
                const index = winners.indexOf(winner);
                delete winners[index];
            }
            console.log('Removed less than level 1', winners);
            let newWinner = getWinners(winners, 1);
            winners = winners.concat(newWinner);
            console.log('New Winner array:', winners);
            returnedArray = winners;
        });
    }
    return returnedArray;
}
// 503418431861948418 should be removed, the other should stay.
// This code returns undefined before the loop even starts. I want this to run AFTER the loop has completed, no matter the amount of entries the array has.
cleanWinners(['209797091457761280','503418431861948418'], message.guild.id)
.then(res => console.log(res));

這是我在控制台中得到的結果：

undefined
Less than 1 level removed [ '209797091457761280', '503418431861948418' ]
New winners array [ '209797091457761280', '503418431861948418' ]
Less than 1 level removed [ '209797091457761280', <1 empty item> ]
New winners array [ '209797091457761280', <1 empty item> ]```

Answer 1

問題是 XP.findOne function 是異步的，你向它傳遞了一個回調。 所以它不會等到一切都完成后再return returnedArray 。 您應該在 findOne 上使用 await await XP.findOne(...然后您的代碼將按照您期望的方式工作。

如果您想查看https://developer.mozilla.org/en-US/docs/Learn/JavaScript/Asynchronous/Async_await ，這里有一個很棒的異步等待 mozilla 鏈接