How do I use async in node.js

Question

I have created a code in node.js, which use mongoDB. Everything is working fine, but when I use nested loop, it break and don't return the proper values.

DB1.find({},function(error,fetchAllDB1) {
    var mainArray = new Array();
    for (var i in fetchAllDB1) {
        if(fetchAllDB1[i].name) {
            var array1  = new Array();
            var array2  = new Array();

            array1['name'] = fetchAllDB1[i].name;
            array1['logo'] = fetchAllDB1[i].logo;
            array1['desc'] = fetchAllDB1[i].desc;    

            DB2.find({is_featured:'1', brand_id: fetchAllDB1[i]._id}, function(error,fetchDB2) {
                for (var p in fetchDB2) {
                    var pArr=[];
                    pArr['_id']    = fetchDB2[p]._id;
                    pArr['name']   = fetchDB2[p].name;
                    pArr['sku']    = fetchDB2[p].sku;
                    pArr['description'] = fetchDB2[p].description;

                    console.log('>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>');
                    console.log(pArr);
                    console.log('<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<');
                    array2[p]=pArr;
                } 
                array1['pp']= array2;
            });
            mainArray[i]=array1;
        }
    }
    console.log('######### LAST #########');
    console.log(mainArray);
    console.log('######### LAST #########');
});

In my console message its showing

 console.log('######### LAST #########');

  All Values

  console.log('######### LAST #########');

After that

console.log('>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>');
All Values;
console.log('<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<');

I want to use async in my code, so that all data fetching occur as below:

mainarray =
[
    [array1] = All Values
    [array2] = [0]
               [1]  
]

mainarray =
[
    [array1] = All Values
    [array2] = [0]
               [1]  
]

How does your code even work? I mean `for (var i in fetchAllDB1)` the `i` should be an object but you use it as an integer? — Stan, Aug 23 '16 at 14:49
adding to @Stan 's comment, what is DB2? the collection or the actual db? If it's the DB, what are you querying for? I don't see in the docs that the db has a find method... — Pedro Otero, Aug 23 '16 at 14:54
Anyway, what you are looking for is this library, there you can make asynchronous actions in loops. https://caolan.github.io/async/ — Stan, Aug 23 '16 at 14:58
Don't use async (advice). Use promises instead. Your code will be much more clear and easily debugable. It's true that it will require a bit effort to handle it if you didn't use promises before, but probably not much than what will be required to fix and maintain async based approach. — bitifet, Aug 23 '16 at 15:57

score -1 · Answer 1 · edited May 23 '17 at 10:28

-1

Your code is very confusing and overly complicated.

What you should do is (psuedo code):

DB1.find(foo) // returns one result
    .then(getFromDB2);

function getFromDB2 (res){
    return DB2.find(res); // make a query here with mongo drivers `$in`
}

Mongodb : $in operator vs lot of single queries

edited May 23 '17 at 10:28

Community

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answered Aug 23 '16 at 15:45

Patrick Motard

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How do I use async in node.js

1 Answers1