坚持在forEach循环中的javascript承诺

Question

让我们考虑维护各种类型得分的字典，比如{a：0，b：0，c：0}我有3个mongo db A，B和C集合表演，

A.find({<key>:<value>})
.then(res=>{
    //gets an array of the response <RES1>
})

现在对于数组中的每个元素，我都是这样做的

B.find({_id : <RES1[i].some_id>})
.then(res=>{
    //again a new array of responses <RES2>
})

最后，我得到了需要在集合C中更新分数的类型，我再次在循环中查询

C.find({_id : RES2[i].some_key})....

我不确定我错在哪里！嵌入代码如下

getScoreForEachTypes() {
    User.find({ company: req.params.company })
        .then(users => {
            var responsesOfAllUsersArr = [];
            usersForThisCompany = users;
            users.forEach(user => {
                responsesOfAllUsersArr.push(Response.find({ email: user.email }));
            });
            return Promise.all(responsesOfAllUsersArr);
        })
        .then(responsesOfAllUsersArr => {
            var data = { _E: 0, _M: 0, _A: 0, _Q: 0, _E: 0 };
            responsesOfAllUsersArr.forEach(el => {
                el.forEach(_el => {
                    var j = getQuestionType(_el.questionId);
                    data[j] += _el.responseChoice;
                })
            })
        });
}

function getQuestionType(qid) {
    return Question.findOne({ _id: qid })
        .then(el => {
            return el.quesType;
        })
        .catch(err => {
            console.log("err while fetching type of question", err);
        });
}

Answer 1

最后我解决了我的问题。我只需要维护一个数组。

function a(){
    var data = { PE: 0, RM: 0, BA: 0, AIQ: 0, FE: 0 };
    var responsesAll = [];
    var xyz = [];
    return User.find({ company: req.params.company })
        .then(users => {
            var responsesOfAllUsersArr = [];
            usersForThisCompany = users;
            users.forEach(user => {
            responsesOfAllUsersArr.push(Response.find({ email: user.email 
        }));
        });
    return Promise.all(responsesOfAllUsersArr);
    })
    .then(responsesOfAllUsersArr => {
        responsesOfAllUsersArr.forEach(el => {
            el.forEach(_el => {
                responsesAll.push(_el);
                xyz.push(getQuestionType(_el.questionId))
            })
        })
        return Promise.all(xyz);
    }).then(xyz => {
        for (var i = 0; i < xyz.length; i++) {
            data[xyz[i].quesType] += parseInt(responsesAll[i].responseChoice);
        }
        return data;
    })
}