MYSQL和nodejs Async Issue中的嵌套查询

Question

我想为另一个查询的每一行执行一个sql查询，所以我写了一个函数

export function getLocations(req, res) {
let appData = [];
const database = new Database();
database.query('select * from districts')
    .then(rows => {
        rows.forEach(row => {
            const new_database = new Database();
            new_database.query(`SELECT locations.id,locations.name
            FROM locations 
            .....
            WHERE locations.district=?`, row.id)
                .then(sub_rows => {
                    let district = row;
                    district["locations"] = sub_rows;
                    appData.push(district);
                    new_database.close();
                }, err => {
                    return new_database.close().then(() => { throw err; })
                })
                .catch(err => {
                    console.log(err);
                    res.status(500).json("Database Error");
                })
    }, err => {
        return database.close().then(() => { throw err; })
    }).then(result => {
        //final result
        res.status(200).json(appData);
        database.close()
    })
    .catch(err => {
        console.log(err);
        res.status(500).json("Database Error");
    })
}

我的数据库类是

const mysql = require('mysql');

class Database {
constructor() {
    this.connection = mysql.createConnection({
        host: process.env.DB_HOST,
        user: process.env.DB_USERNAME,
        password: process.env.DB_PASSWORD,
        database: process.env.DB_NAME,
        port: 3306,
        debug: false,
        multipleStatements: false
    });
}
query(sql, args) {
    return new Promise((resolve, reject) => {
        this.connection.query(sql, args, (err, rows) => {
            if (err)
                return reject(err);
            resolve(rows);
        });
    });
}
close() {
    return new Promise((resolve, reject) => {
        this.connection.end(err => {
            if (err)
                return reject(err);
            resolve();
        });
    });
}
}
module.exports = Database;

然而，这在执行时继续并打印最终结果（空数组）而不等待执行子查询。我在row和new_database.query上尝试了异步等待，但它仍然没有等待子查询执行完成。我能在这做什么？

Answer 1

我已经改进了sub-queries的逻辑并使用Promise.all()重新编写它们：

export function getLocations(req, res) {
    let appData = [];
    const database = new Database();
    database.query('select * from districts')
        .then(rows => {

            // create promise list for all sub-queries
            const promise_list = rows.map(row => new Promise((resolve, reject) => {
                const new_database = new Database();
                return new_database.query(`SELECT locations.id,locations.name FROM locations .... WHERE locations.district=?`, row.id)
                    .then(sub_rows => {
                        let district = row;
                        district["locations"] = sub_rows;
                        appData.push(district);
                        new_database.close();
                        resolve(district);
                    }, err => {
                        return new_database.close().then(() => { throw err; })
                    });
            })
            );

            // execute all sub-queries in parallel and wait for them
            Promise.all(promise_list).then(result => {
                res.status(200).json(appData);
                database.close()
            }).catch(err => {
                console.log(err);
                res.status(500).json("Database Error");
            });

        });
}

PS：你可以将appData声明为const，你只是将数据推入其中。没有重新分配它。

Answer 2

使用JOIN尝试获取您所追求的结果。可用的功能取决于MySQL版本，因此您可能需要GROUP_CONCAT而不是JSON_ARRAYAGG

SELECT districts.*, JSON_ARRAYAGG(locations)
FROM districts
LEFT JOIN locations
   ON locations.district=districts.id
GROUP BY districts.id

Javascript连接具有可怕的性能，所以让数据库做它的设计目的。