在Nodejs中从mySQL获取数据时如何从异步/等待函数中获取返回值

Question

我从DB中获取一些exchange数据，然后提取不同交换的名称，然后再转到MYSQL查询中，从不同的表中获取数据。

我面临的问题是异步等待不返回值，而只是返回允诺{ }。

下面是我正在尝试的代码，想知道我哪里出错了。

//Function that fetches the exchanges from DB

const getExchange = () => {
        return new Promise((resolve, reject) => {
            db.connection.query(`
            SELECT  *
            FROM,
            (
            SELECT
                exchange,
                COUNT(pair) as noOfMarkets
                FROM ticker_data

        ) as t
            `, (err, resp) => {
                if (!err) {
                    resolve(resp)
                } else {
                    reject(err)
                }
            })
        })
    }


// push unique exchanges to an array.
const getExchangesData = async () => {
    const allExchanges = await getExchanges();

    let exchanges = []
    allExchanges.forEach(item => {
        let exchange = {
            exchange: item.exchange
        }
        exchanges.push(exchange)
    })
    return await exchanges
}


// mapping through an array of exchanges and passing to DB query to get data from the DB.

const getSingleExchange = async () => {
    const exchanges = await getExchangesData()
    await Promise.all(exchanges.map(async (item) => {
        db.connection.query(`
        SELECT
       exchange_rank,
       name
       volume24hUSD
       (
            SELECT
            volume24hUSD as tradingVolYesterday
            FROM exchanges
            WHERE name = '${item.exchange}'
            AND createdAt >= now() -interval 1 day
            AND createdAt <  now() -interval 1 day + interval 120 second
            LIMIT 1
       ) volumeDay1
FROM exchanges
WHERE name = '${item.exchange}'
        `, (err, resp) => {
            if (!err) {
                console.log(resp) // getting all the values

                let volData = {
                     name: resp[0].name,
                     exchange_rank: resp[0].exchange_rank,
                     icon: resp[0].icon
                 }
                 return volData 
            }
        })
    }))
}


const data = getSingleExchange()
console.log(data) // returning Promise { <pending> } 

编辑

在作出答案所建议的改变后，我仍然有一个问题：

//Function that fetches the exchanges from DB

const getExchange = () => {
        return new Promise((resolve, reject) => {
            db.connection.query(`
            SELECT  *
            FROM,
            (
            SELECT
                exchange,
                COUNT(pair) as noOfMarkets
                FROM ticker_data

        ) as t
            `, (err, resp) => {
                if (!err) {
                    resolve(resp)
                } else {
                    reject(err)
                }
            })
        })
    }


// push unique exchanges to an array.
const getExchangesData = async () => {
    const allExchanges = await getExchanges();

    let exchanges = []
    allExchanges.forEach(item => {
        let exchange = {
            exchange: item.exchange
        }
        exchanges.push(exchange)
    })
    return await exchanges
}


    // mapping through an array of exchanges and passing to DB query to get data from the DB.

const getSingleExchange = async () => {
    const exchanges = await getExchangesData()
    await Promise.all(exchanges.map((item) => {
        return new Promise((resolve, reject) => {
            db.connection.query(`...`, (err, resp) => {
                if (!err) {
                    resolve(resp)
                } else {
                    reject(err)
                }
            }).then(resp => {
                console.log(resp)
                let volData = {
                    name: resp[0].name,
                    exchange_rank: resp[0].exchange_rank,
                    icon: resp[0].icon
                }
                return volData
            })
        })
    }))

}

getSingleExchange().then(data => {
    console.log(data)
});

我现在得到了这个错误：

(节点:30583) UnhandledPromiseRejectionWarning: TypeError：db.connection.query(...).then不是函数 准承诺(/getExchanges.js:217:16) 新承诺() Promise.all.exchanges.map (/getExchanges.js:145:16) Array.map () getSingleExchange (/getExchanges.js:144:33)

Answer 1

这一部分的主要问题是：

await Promise.all(exchanges.map(async (item) => {

map回调没有返回任何内容，也没有await，因此使用async是没有意义的。

相反，删除async

await Promise.all(exchanges.map((item) => {

..。并在回调函数中返回一个承诺，就像在第一个函数中所做的那样：

    return new Promise((resolve, reject) => {
        db.connection.query(`...`), (err, resp) => {
            if (!err) {
                resolve(resp)
            } else {
                reject(err)
            }
        })
    }).then(resp => {
        console.log(resp)
        let volData = {
             name: resp[0].name,
             exchange_rank: resp[0].exchange_rank,
             icon: resp[0].icon
         }
         return volData 
    });

编写一个促进query的泛型函数将使您受益，这样您就不必为所需的每一个查询执行new Promise-thing。

最后，不能期望同步地获得异步结果：async函数不同步返回异步结果，而是返回承诺。因此，您的最后一行(主要代码)仍应等待。所以要么这样做：

(async () => {
    const data = await getSingleExchange()
    console.log(data)
})(); // immediately executing (async) function expression

或者：

getSingleExchange().then(data => {
    console.log(data)
});

注意:在第二个函数中执行return await exchanges是没有意义的(exchanges不是一个承诺)，所以您可以只做return exchanges。