如何在mysql中用一行命令在兩個表中插入資料？

首先，我看到了一些接近這個問題的答案，但它不起作用。這就是為什么我創建了一個新問題，這里是規范：

我創建了兩個表：

CREATE TABLE car_info (
    id int PRIMARY KEY AUTO_INCREMENT NOT NULL, 
    mileage int, 
    make varchar(100), 
    model varchar(100), 
    fuel varchar(100), 
    gear varchar(100), 
    offertype varchar(100)
);

CREATE TABLE sales_info (
    id int PRIMARY KEY AUTO_INCREMENT NOT NULL, 
    FOREIGN KEY(id) REFERENCES car_info(id), 
    price int, 
    hp int, 
    year int
);

我加入了這兩個：

select * 
from (SELECT * 
        FROM car_info
     ) as a 
    left outer join (SELECT * 
                    FROM sales_info
                    ) as b on a.id = b.id 
union 
select * 
from (SELECT * 
        FROM car_info
      ) as a 
    right outer join (SELECT * 
                    FROM sales_info
                    ) as b on a.id = b.id;

如您所見， id 是這里的公共變數。這里是 car_info 表：

 ----------- -------------- ------ ----- --------- ---------------- 
| Field     | Type         | Null | Key | Default | Extra          |
 ----------- -------------- ------ ----- --------- ---------------- 
| id        | int          | NO   | PRI | NULL    | auto_increment |
| mileage   | int          | YES  |     | NULL    |                |
| make      | varchar(100) | YES  |     | NULL    |                |
| model     | varchar(100) | YES  |     | NULL    |                |
| fuel      | varchar(100) | YES  |     | NULL    |                |
| gear      | varchar(100) | YES  |     | NULL    |                |
| offertype | varchar(100) | YES  |     | NULL    |                |
 ----------- -------------- ------ ----- --------- ---------------- 

這里是 sales_info 表：

 ------- ------ ------ ----- --------- ---------------- 
| Field | Type | Null | Key | Default | Extra          |
 ------- ------ ------ ----- --------- ---------------- 
| id    | int  | NO   | PRI | NULL    | auto_increment |
| price | int  | YES  |     | NULL    |                |
| hp    | int  | YES  |     | NULL    |                |
| year  | int  | YES  |     | NULL    |                |
 ------- ------ ------ ----- --------- ---------------- 

這是我通常用來在單個表中匯入 SQL 資料的代碼：

app.post('/save', (req, res) => {
    let data = { mileage: req.body.mileage, make: req.body.make, model: req.body.model, fuel: req.body.fuel, gear: req.body.gear, offertype: req.body.offertype };
    let sql = "INSERT INTO car_info SET ?";
    let query = connection.query(sql, data, (err, result) => {
        if (err) throw err;
        res.redirect('/');
    });
});

當我使用此代碼并從表單中呼叫它時，它可以完美運行。但是，我想要做的是，我想從表單中呼叫此代碼并將兩個表匯入到相關資訊中，例如：

app.post('/save', (req, res) => {
    let data = { mileage: req.body.mileage, make: req.body.make, model: req.body.model, fuel: req.body.fuel, gear: req.body.gear, offertype: req.body.offertype, price: req.body.price, hp: req.body.hp, year: req.body.year };
    let sql = "**car_info and sales_info tables code here**";
    let query = connection.query(sql, data, (err, result) => {
        if (err) throw err;
        res.redirect('/');
    });
});

最后，我嘗試了類似的方法：

let sql = `INSERT INTO car_info 
                    (mileage, make, model, fuel, gear, offertype) 
            VALUES('${req.body.mileage}', '${req.body.make}', 
                    '${req.body.model}', '${req.body.fuel}', 
                    '${req.body.gear}', '${req.body.offertype}');
           INSERT INTO sales_info
                    (price, hp, year) 
            VALUES('${req.body.price}', '${req.body.hp}', 
                    '${req.body.year}');`

但這里的問題是，它只在 car_info 中插入資料，而不在 sales_info 中插入資料，盡管我在相關欄位中輸入了一些值。

我希望這個問題很清楚，我怎樣才能做到這一點？

uj5u.com熱心網友回復：

由于您的兩個表具有不同的列，我建議您在兩個不同的變數中創建兩個查詢，然后相應地插入從req.body表中收集的資料。

app.post('/save', (req, res) => {
    const { milage, make, model, fuel, gear, offertype, price, hp, year } = req.body //Destructured object
    const car_info_query = `INSERT into car_info VALUES (?,?,?,?,?,?)`
    const sales_info_query = `INSERT into sales_info VALUES (?, ?, ?)`
    
    let write_car_info = connection.query(car_info_query, [milage, make, model, fuel, gear, offertype] , (err, result) => {
        if (err) throw err;
    });
    let write_sales_info = connection.query(sales_info_query, [price, hp, year], (err, result) => {
        if (err) throw err
    })
    //Insert into as many tables as you want
    res.redirect('/');
});

uj5u.com熱心網友回復：

按照@jkalandarov 的回答，我解決了這個問題。這里的主要問題是 id 是外鍵的 sales_info 的關系。但最后，它解決了:)

這是我的最終代碼：

app.post('/save', (req, res) => {
    let car_info_data = { mileage: req.body.mileage, make: req.body.make, model: req.body.model, fuel: req.body.fuel, gear: req.body.gear, offertype: req.body.offertype };
    let sales_info_data = { price: req.body.price, hp: req.body.hp, year: req.body.year };

    const car_info_query = 'INSERT into car_info SET ? ';
    const sales_info_query = 'INSERT into sales_info SET ?';

    let write_car_info = connection.query(car_info_query, car_info_data, (err, result) => {
        if (err) throw err;
    });
    let write_sales_info = connection.query(sales_info_query, sales_info_data, (err, result) => {
        if (err) throw err;
    })
    //Insert into as many tables as you want
    res.redirect('/');
});

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