我無法在.then（）中返回對象

Question

我創建了一個id對象，但是我不能在for或下一個.then（）中訪問它，我將det設置為全局變量，有人知道它在哪里嗎？

.then(function(idHome){

    home_id = idHome;

    var detname_img = [];
    var sqlEdit = "INSERT INTO images_det SET ?";

    for(var k = 0; k < object.det_img.length; k++){

        detname_img.push({
            name_img : object.det_img[k].name_img,
        });

        connection.query(sqlEdit,detname_img[k]);
    }

    // console.log(detname_img);

    for(var i = 0; i < detname_img.length; i++){

        var getDet = ("SELECT det_id from images_det where name_img = '" + detname_img[i].name_img + " ' order by 1 desc limit 1");

        connection.query(getDet, function(erro, result){

            for(var k = 0; k < result.length; k++){

                det.push({
                    det_id : result[k].det_id
                });
                console.log(chalk.blue(det[0].det_id));
                return det;
            }
        });
    }        

});

Answer 1

假設您使用此節點mysql包，則可以從insert語句的結果中獲取插入的ID。

由於該軟件包似乎沒有意識到promise的存在，因此您可以使用將返回promise的包裝器將api除塵：

//turn an insert query with callback function 
//  into a function returning a promise
const insertAsPromise = connection => args => 
  new Promise(
    (resolve,reject)=>
      //call connection query with variable amount of argument(s)
      //  using Function.prototype.apply
      connection.query.apply(connection,args.concat(
        //add the callback to the list of arguments, callback
        //  is the last argument
        (error, results, fields) =>
          (error)
            ? reject(error)
            : resolve([results,fields])
      ))
  );


const sqlEdit = "INSERT INTO images_det SET ?";
Promise.all(
  object.det_img.map(
    image=>//map magical/global object.det_img to promise
      insertAsPromise(connection)([sqlEdit,image.name_img])
      .then(//when promise resolves get the inserted id
        ([results,fields])=>
          results.insertId
      )
  )
)
.then(
  results=>console.log(results)//should log an array of id's
)
.catch(
  err=>console.error("something went wrong:",err)
)