Javascript - 待定承诺

Question

我正在学习承诺的工作，我想从表中选择数据并将其添加到数组中，然后将该数组发送到下一个。

我这样做，但它返回[Promise {<pending>}]：

    var p1 = new Promise(function(resolve, reject) {
      var stamp = req.params.stamp;
      var part = [];
      part.push(request.query("SELECT u_order, u_familia, u_part, u_type FROM u_part (nolock) where u_order <'92' and u_order <> '100'"));
      resolve(part);
      // or
      // reject ("Error!");
    });
    p1.then(function(value){
      console.log(value);
    }, function(reason){
      console.log(reason);
    }); 

我认为我必须等待承诺，但如何？

更新

现在我想迭代数组部分并选择另一个数组。之后，我必须使用此结构创建一个新数组：

-PARTS
      - part
            - u_order
            - u_familia
            - u_part
            - u_type
            - articles (article from each part)

我尝试了这个，但无法访问数组文章

var p1 = new Promise(function(resolve, reject) {
      var stamp = req.params.stamp;
      request.query("SELECT u_order, u_familia, u_part, u_type FROM u_part (nolock) where u_order <'92' and u_order <> '100'").then((data)=>resolve(data));
      // or
      // reject ("Error!");
    });

    p1.then(function(value){
      var stamp = req.params.stamp;
      console.log(value.length);
      for(var i= 0; i<value.length; i++)
      {
        console.log(value[i]);
        request.query("SELECT st.u_posic, sc.ref, sc.qtt, sc.design FROM st INNER JOIN sc ON st.ref = sc.ref where sc.ststamp ='"+stamp+"' and st.u_posic = '"+value[i].u_order+"'").then((data)=>resolve(data));
      }
    }, function(reason){
      console.log(reason);
    });
    p1.then(function(part, articles){
      var parts = [];
      console.log("PART: " +part.length);
      for(var j= 0; j<part.length; j++)
      {
        console.log(part[j].u_order);
        console.log(part[j].u_familia);
        console.log(part[j].u_part);
        console.log(part[j].u_type);
        console.log(articles[j]);
      };
    });

谢谢

Answer 1

您可以执行以下操作：

var stamp = req.params.stamp;
var p1 = request.query("SELECT u_order, u_familia, u_part, u_type FROM u_part (nolock) where u_order <'92' and u_order <> '100'");
p1.then((value) => {
    console.log(value.length);
    var promise_arr = [];
    for(var i= 0; i<value.length; i++)
    {
        console.log(value[i]);
        promise_arr.push(request.query("SELECT st.u_posic, sc.ref, sc.qtt, sc.design FROM st INNER JOIN sc ON st.ref = sc.ref where sc.ststamp ='"+stamp+"' and st.u_posic = '"+value[i].u_order+"'"));
    }
    //Since request.query returns a promise we need to resolve all of the promises using Promise.all
    return Promise.all(promise_arr);
}, (reason) => {
    console.log(reason);
}).then((part) => { //This is where you receive result form Promise.all
    var parts = [];
    console.log("PART: " +part.length);
    for(var j= 0; j<part.length; j++)
    {
        console.log(part[j].u_order);
        console.log(part[j].u_familia);
        console.log(part[j].u_part);
        console.log(part[j].u_type);
        console.log(articles[j]);
    };
})