如何从MySQL查询返回json到GraphQL？

我是nodejs和qraphql的新手，我正在尝试对graphql进行简单的查询 . 在对MySQL数据库进行查询之后，我的问题是我不知道如何正确地将数据发送到graphql以查看结果 .

我必须尝试使用解决方案，但它不起作用 . 我使用json从mySQL返回结果，但它不起作用 . 为什么我在Graphiql中看不到结果？我该如何回归？

这是我进行查询并返回结果的函数：

let getPlayer = (args) => {
    let id = args.id;
    console.log("id: " + id);
    let myPromise = new Promise((resolve, reject) => {
        const connection = mysql.createConnection(config.ddbb_connection);
        connection.connect();
        connection.query("select json_object('id', id) as player from tbl003_player where id = " + id, 
                        function (error, results, fields) {
            if (!error){
                if (results.length > 0){
                    resolve(results[0]);
                    console.log("resultado: " + results[0]);
                }else{
                    reject(new Error("No se ha encontrado ninguna jugadora con el ID: " + id));
                }
            }else{
                reject(new Error("Se ha producido un error de acceso a BBDD"));
            }             
          });
        connection.end();        
    });
    console.log("Salgo de la consulta");
    myPromise.then((resolve) => {
        console.log("resolve: " + JSON.stringify(resolve));
        return resolve;
    },(error) => {
        console.log(error);
        return error;
    });
};

Edit I:

如果我为JSON.parse更改JSON.stringify，我在控制台中遇到此错误：

(node:31898) UnhandledPromiseRejectionWarning: SyntaxError: Unexpected token o in JSON at position 1
    at JSON.parse (<anonymous>)
    at myPromise.then (/home/josecarlos/Workspace/graph-ql/primer-server-express/routes-api.js:68:21)
    at <anonymous>
    at process._tickCallback (internal/process/next_tick.js:188:7)
(node:31898) UnhandledPromiseRejectionWarning: Unhandled promise rejection. This error originated either by throwing insideof an async function without a catch block, or by rejecting a promise which was not handled with .catch(). (rejection id: 1)
(node:31898) [DEP0018] DeprecationWarning: Unhandled promise rejections are deprecated. In the future, promise rejections that are not handled will terminate the Node.js process with a non-zero exit code.

Edit II:

我已经修复了错误，当我返回JSON.parse（解析）但它不起作用:(我仍然在GraphiQL中收到此错误：

{
  "data": {
    "player": null
  }
}

我管理myPromise的代码现在就是这个......

myPromise.then((resolve) => {
    console.log("resolve: " + JSON.stringify(resolve));
    data = JSON.parse(resolve);
    return data;
},(error) => {
    console.log(error);
    return error;
}).catch(() => {
    console.log("Entro dentro del catch");
});

Edit III:

我们用JSON.stringify（resolve）得到了这个字符串{"player"："{" id \ ": 11}"} . 我想我只需要回复{\ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _我怎样才能做到这一点？知道如何将json返回到GraphiQL吗？

Edit IV:

我修改了我的代码只返回带有返回解析的json（结果[0] .player）并且它不起作用！

这是我的实际代码：

let getPlayer = (args) => {
    let id = args.id;
    console.log("id: " + id);
    let myPromise = new Promise((resolve, reject) => {
        const connection = mysql.createConnection(config.ddbb_connection);
        connection.connect();
        connection.query("select json_object('id', id) as player from tbl003_player where id = " + id, 
                        function (error, results, fields) {
            if (!error){
                if (results.length > 0){
                    resolve(results[0].player);
                    console.log("resultado: " + results[0].player);
                }else{
                    reject(new Error("No se ha encontrado ninguna jugadora con el ID: " + id));
                }
            }else{
                reject(new Error("Se ha producido un error de acceso a BBDD"));
            }             
          });
        connection.end();        
    });
    console.log("Salgo de la consulta");
    myPromise.then((resolve) => {
        console.log("resolve: " + JSON.stringify(resolve));
        return JSON.parse(resolve);
    },(error) => {
        console.log(error);
        return error;
    }).catch(() => {
        console.log("Entro dentro del catch");
    });
};