在javascript中按顺序履行承诺[重复]

Question

我一直在尝试通过寻找工作时间最短的人来使用承诺来履行轮班工作。为此，我使用聚合将每个人的可用性文档与包含他们上次工作时间的记录文档链接起来。似乎它跳过了与承诺有关的整个代码部分，因为它将 selectedPeopleList 打印为空数组。这是我的代码：

var selectedPeopleList = [];

var sequentially = function(shifts) {

var p = Promise.resolve();

shifts.forEach(function (){
   p=p.then( function() { return collection.aggregate([
       {
          $lookup: 
            {
              from: "personRecord",
              localField: "ATTU_ID",
              foreignField: "ATTU_ID",
              as: "record"
            } 
        },

        {
          $match : { 'Available[]' : { $elemMatch : { $eq : shift.value } }, "record.ATTU_ID": { $nin : _.map(selectedPeopleList, 'ATTU_ID') } }
        },

        { 
          $sort : { "record.lastShift" : 1 }
        }
          ]).toArray(function(err, docs){ 
            assert.equal(err, null);

          }).then(function (result) {
            if(docs && docs.length) {
              selectedPeopleList.push({ ATTU_ID : docs[0].ATTU_ID, Name: docs[0].Name });
              console.log(docs[0]);
            }
          });
        });            
  })
  return p;
};
console.log(selectedPeopleList);

Answer 1

Promise 不会使异步代码同步

把最后一行改成

p.then(function() {
    console.log(selectedPeopleList);
});

另外，您不需要在 forEach 中返回任何内容，因为根本不使用返回值

Answer 2

您的代码中有不少小错误。

确保正确遵循异步代码流。我试图解决我的 sn-p 中的所有问题。

var selectedPeopleList = [];

var sequentially = function(shifts) {

var p = Promise.resolve();
//use promise all to merge all promises in to 1. and use map to return the promises.
Promise.all(shifts.map(function (){
   p=p.then( function() { return collection.aggregate([
       {
          $lookup: 
            {
              from: "personRecord",
              localField: "ATTU_ID",
              foreignField: "ATTU_ID",
              as: "record"
            } 
        },

        {
          $match : { 'Available[]' : { $elemMatch : { $eq : shift.value } }, "record.ATTU_ID": { $nin : _.map(selectedPeopleList, 'ATTU_ID') } }
        },

        { 
          $sort : { "record.lastShift" : 1 }
        }
          ]).toArray(function(err, docs){ 
            assert.equal(err, null);
          })
   })//make sure you finish then! before you start a new one.
   
   .then(function (result) {
            console.log('our results', result);
            if(result && result.length) { //use result variable!
              selectedPeopleList.push({ ATTU_ID : result[0].ATTU_ID, Name: result[0].Name });
              
            }
         
        });            
  })

  return p;
})
)
.then(function(){ //lets await all the promises first. and then read the list.
  
  console.log(selectedPeopleList);
  });