交叉引用 JSON 和 MySQL？

Question

我为一个不和谐的机器人编写了一个“购买”功能，它将项目的id 插入到 mysql 中的相应表中。

购买示例：

const join = args.join(" ")
let item = JSON.parse(fs.readFileSync('commands/economy/resources/items.json', "utf8"));


//Horses
const Horse1 = item.Horse1


        //Buy Horse1
        if (join === Horse1.name) {

            let itemName = Horse1.name
            let itemLuck = Horse1.luck
            let itemId = Horse1.ID.toString()
            let itemPrice = Horse1.price.toString()

            //Account Check
            if (rows.length < 1) { return message.channel.send("You do not have an account with The Iron Bank. Use ``;start`` to open one.") };
            //Item Check
            if (rows[0].horse === itemId) { return message.channel.send(`❌ | You already have a ${itemName}!`) }
            //Money Check
            if (rows[0].bank < itemPrice) { return message.channel.send(`You do not have enough coins in your account to buy **${itemName}**. It costs ${itemPrice} coins. You have ${rows[0].bank} coins`) };


            //DB Update
            con.query(`UPDATE economy SET horse = '${itemId}', horseluck = '${itemLuck}', bank = bank-${itemPrice} WHERE user = ${message.author.id}`)

            //Bought
            message.channel.send(`✅ | Purchased ${itemName} for ${itemPrice} coins.`)
        };

长话短说，我需要从id: 中提取name: 并与rows[0].horse / rows[0].weapon 交叉引用它

这是我的 mySQL 选择查询中的代码：

  //Direct to start if not in DB
    if (rows.length < 1) {

        return message.channel.send("You do not have an account with The Iron Bank. Use ``;start`` to open one.")

    };

    let horse = rows[0].horse;
    let weapon = rows[0].weapon;
    let armour = rows[0].armour;
    let luck = (rows[0].horseluck + rows[0].weaponluck + rows[0].armourluck)
    console.log()

    return message.channel.send(`**Horse**: ${horse}\n**Weapon**: ${weapon ? weapon : "None"}\n**Armour**: ${armour ? armour : "None"}\n**Luck Bonus**: ${luck ? "+"+luck+"%" : "None"}`);

但上面打印出马的 ID 而不是 message.channel.send 中的名称。我如何将 MySQL 中的 ID 交叉引用到 JSON 中的 ID 并在 message.channnel.send 中打印出名称？

Example of my JSON file:

    {
"Horse1":{
        "ID": 1,
        "name": "Dornish Sand Steed",
        "description": "Splendid beasts",
        "type": "Horses",
        "price": "100000",
        "luck": "5.0"
    },
 "Horse2":{
        "ID": 2,
        "name": "Arabian",
        "description": "Preferred war mounts",
        "type": "Horses",
        "price": "50000",
        "luck": "2.0"
    },
"Weapon1":{
        "ID": 1,
        "name": "Longclaw",
        "description": "Valyrian steel sword that was the ancestral weapon of House Mormont",
        "type": "Weapons",
        "price": "100000",
        "luck": "5.0"
    },
"Weapon2":{
        "ID": 2,
        "name": "Oathkeeper",
        "description": "Valyrian steel sword made from House Starks greatsword",
        "type": "Weapons",
        "price": "50000",
        "luck": "2.0"
    },
}

（我知道这段代码中的 SQL 查询容易被 SQL 注入，这只是为了在我开始使用 mysql.format 之前进行测试）

Answer 1

设法解决了这个问题：

使用Object.values(item) 将 JSON 对象值转换为数组，然后在将 rows[0] 值转换为 INT 后使用 ID 索引数组，如下所示。如果该值未定义（在这种情况下为 null，因为它不能具有除 JSON 中指定的 ID 之外的任何其他值）视为“无”。

    let horse = rows[0].horse;
    let weapon = rows[0].weapon;
    let armour = rows[0].armour;
    let luck = (rows[0].horseluck + rows[0].weaponluck + rows[0].armourluck)

    function getItemName(id){
        return (id == undefined)? "None" : Object.values(item)[id - 1].name;
    }

    let horseName = getItemName(parseInt(horse))
    let weaponName = getItemName(parseInt(weapon) + 5)
    let armourName = getItemName(parseInt(armour) + 10)

    //5 items of each type


    return message.channel.send(`**Horse**: ${horseName}\n**Weapon**: ${weaponName}\n**Armour**: ${armourName}\n**Luck Bonus**: ${luck ? "+"+luck+"%" : "None"}`);