如何在 JS 的 Promise 中使用第二个查询？使用 mysql 数据库

Question

基本上我有一个 createorder 需要将订单添加到数据库中，并将订单中的所有产品添加到 Orderline 额外表中。

一直在尝试不同的方法来使第二个查询工作，但唯一插入到数据库中的是来自第一个查询的数据。

第二个查询也需要使用第一个查询的插入顺序中的id。因为数据库对订单的 id 使用自动增量。

CreateOrder: (req) => {
        return new Promise((resolve, reject) => {
            const sqlInsert = 'INSERT INTO web3.orders (Created, firstName, lastName, street, number, postalCode, city, telephone, email, totalPrice) VALUES (?,?,?,?,?,?,?,?,?,?);';
            const sqlInsert2 = 'INSERT INTO web3.orderline (orderId,productId,qty,price) VALUES (?,?,?,?);';

            db.query(sqlInsert, [
                req.body.Created,
                req.body.firstName,
                req.body.lastName,
                req.body.street,
                req.body.number,
                req.body.postalCode,
                req.body.city,
                req.body.telephone,
                req.body.email,
                req.body.totalPrice],
                (err, result, fields) => {
                    if (err) {
                        reject(err);
                    }
                    else {
                        resolve(result);

                    }
                });
                db.query(sqlInsert2,
                    [
                        result.insertId,
                        req.body.producten.id,
                        req.body.producten.cartQuantity,
                        req.body.producten.price], (err, result, fields) => {
                            if (err) {
                                reject(err);
                            } else {
                                resolve(result);
                            }
                        });
                    });
                },
            }

Answer 1

您需要两个承诺：每个查询一个。只有在第一个解决后，才应创建第二个。

一个很好的方法是promisify db.query 到一个新函数中，然后只使用那个函数。所以先定义这个：

const promiseQuery = (sql, params) => new Promise((resolve, reject) =>
    db.query(sql, params, (err, result, fields) => err ? reject(err) : resolve(result))
);

其次，您可以将CreateOrder 设为async 函数，以便您可以使用await：

async CreateOrder: (req) => {
    const sqlInsert = 'INSERT INTO web3.orders (Created, firstName, lastName, street, number, postalCode, city, telephone, email, totalPrice) VALUES (?,?,?,?,?,?,?,?,?,?);';
    const sqlInsert2 = 'INSERT INTO web3.orderline (orderId,productId,qty,price) VALUES (?,?,?,?);';
    let result = await promiseQuery(sqlInsert, [
        req.body.Created,
        req.body.firstName,
        req.body.lastName,
        req.body.street,
        req.body.number,
        req.body.postalCode,
        req.body.city,
        req.body.telephone,
        req.body.email,
        req.body.totalPrice
    ]);
    await promiseQuery(sqlInsert2, [
        result.insertId,
        req.body.producten.id,
        req.body.producten.cartQuantity,
        req.body.producten.price
    ]);
}

Answer 2

上面的代码是正确的，同样的流程在这里工作

async function CreateOrder(req) {
    const sqlInsert = 'INSERT INTO legs (leg) VALUES (?);';
    const sqlInsert2 = 'INSERT INTO destinations (destination,area) VALUES (?,?);';
    let result = await promiseQuery(sqlInsert, [
        "lw"
    ]);
  let res=   await promiseQuery(sqlInsert2, [
        "test",
        "test location"
    ]);
    console.log(res)
}

const promiseQuery = (sql, params) => new Promise((resolve, reject) =>
    db.query(sql, params, (err, result, fields) => err ? reject(err) : resolve(result))
);
async function test(){
   await CreateOrder()
}

尝试检查是否有任何键未定义或字段类型与您传递的值不匹配

如果你 req.body.producten 是一个数组你必须为每次迭代做查询

async CreateOrder: (req) => {
    const sqlInsert = 'INSERT INTO web3.orders (Created, firstName, lastName, street, number, postalCode, city, telephone, email, totalPrice) VALUES (?,?,?,?,?,?,?,?,?,?);';
    const sqlInsert2 = 'INSERT INTO web3.orderline (orderId,productId,qty,price) VALUES (?,?,?,?);';
    let result = await promiseQuery(sqlInsert, [
        req.body.Created,
        req.body.firstName,
        req.body.lastName,
        req.body.street,
        req.body.number,
        req.body.postalCode,
        req.body.city,
        req.body.telephone,
        req.body.email,
        req.body.totalPrice
    ]);
    for (let ele of req.body.producten){
        await promiseQuery(sqlInsert2, [
           result.insertId,
           ele.id,
           ele.cartQuantity,
           ele.price
        ]);
    }
}

Answer 3

试试这个：

CreateOrder: (req) => {
        const { Created, firstName, lastName, street, number, postalCode, city, telephone, email, totalPrice } = req.body
        let results = {}
        return new Promise((resolve, reject) => {
            const sqlInsert = 'INSERT INTO web3.orders (Created, firstName, lastName, street, number, postalCode, city, telephone, email, totalPrice) VALUES (?,?,?,?,?,?,?,?,?,?);';
            const sqlInsert2 = 'INSERT INTO web3.orderline (orderId,productId,qty,price) VALUES (?,?,?,?);';

            db.query(sqlInsert, [Created, firstName, lastName, street, number, postalCode, city, telephone, email, totalPrice],
                (err, result, fields) => {
                    console.log(result)
                    results.insertId = result.id
                });
            db.query(sqlInsert2, [ results.insertId, req.body.producten.id, req.body.producten.cartQuantity, req.body.producten.price], (err, result, fields) => {
                            if (err) {
                                reject(err);
                            } else {
                                resolve(result);
                            }
                        });
                    });
                },
            }

我猜在您的初始代码中，您正在解决第一个 db.query 中的承诺，这导致承诺结束。