Mongo DB returns undefined values in an array

I have this code that takes the value of the number of documents on a given day of the week in MongoDB. And as a return request, the "qweek" array is filled.

function dates(current) {
  var week = new Array();

  // Starting Monday not Sunday
  current.setDate((current.getDate() - current.getDay() + 1));

  for (var i = 0; i < 7; i++) {
    var dd = String(current.getDate()).padStart(2, '0');
    var mm = String(current.getMonth() + 1).padStart(2, '0'); //January is 0!
    var yyyy = current.getFullYear();
    var day = dd + '/' + mm + '/' + yyyy;

    week.push(day);

    current.setDate(current.getDate() + 1);
  }

  return week;
}

// Initialize the App Client
const client = stitch.Stitch.initializeDefaultAppClient("app-id");

// Get a MongoDB Service Client
const mongodb = client.getServiceClient(
  stitch.RemoteMongoClient.factory,
  "mongodb-atlas"
);

//projection config
const options = { // Match the shape of RemoteFindOptions.
  limit: 1000, // Return only first ten results.
  projection: { // Return only the `title`, `releaseDate`, and
    day: 1, //   (implicitly) the `_id` fields.
  },
  sort: { // Sort by releaseDate descending (latest first).
    releaseDate: -1,
  },
}

// Get a reference to the travels database
const db = mongodb.db("travels");

function displayCountTravels() {
  var daysweek = dates(new Date());
  var qweek = new Array();

  for (var l = 0; l < daysweek.length; l++) {
    db.collection("details")
      .find({
        "day": daysweek[l]
      }, options)
      .toArray()
      .then(docs => {
        qweek.push(docs.length);
      });
  }

  console.log(qweek);
  console.log(qweek[1]);

  return qweek;
}

In this case, when I make a request in the array console. I get this return:

console.log(qweek);
Log output:[]
0: 0
1: 0
2: 0
3: 2
4: 0
5: 0
6: 0
length: 7
__proto__: Array(0)

Return of command console.log(week);

Return of command "console.log(week);"

But when I try to get the value by the index. The array item is returned with undefined.

console.log(qweek[1]);

Log output:

undefined

Return of command console.log(week[1]);

Return of command "console.log(week[1]);"

I would like to know why the value comes with undefined.

Upvotes: 2

Views: 1659

Answers (1)

ambianBeing
ambianBeing

Reputation: 3529

Essentially this is a case of asynchronous behavior in Javascript. On top of that asynchronous calls are made in a for..loop.

Short explanation: The mongo-db query calls are async in nature and the execution is not going to wait for it finish before reaching the console.log(qweek) which is outside the then block. As a result you'll be getting qweek as empty[] or qweek[1] as undefined.

Couple of ways solve this is Serializing with promises and async/await or using Promise.all(). Would suggest you to read upon them to understand more.

Using async/await: Syntax wise less verbose and easy to comprehend

async function displayCountTravels() {
  var daysweek = dates(new Date());
  var qweek = [];
  try {
    for (var l = 0; l < daysweek.length; l++) {
      /*wait for the promise to resolve before next iteration*/
      var docs = await db
        .collection("details")
        .find({ day: daysweek[l] }, options).toArray();

      qweek.push(docs.length);
    }
  } catch (e) {
    console.error(e);
  }

  console.log(qweek);
  console.log(qweek[1]);

  return qweek;
}

Using Promise.all(...):

function displayCountTravels() {
  var daysweek = dates(new Date());
  var promises = [];

  /*each mongo query is a promise, collect and return*/

  for (var l = 0; l < daysweek.length; l++) {
    promises.push(
      db.collection("details").find({ day: daysweek[l] }, options).toArray()
    );
  }

  return Promise.all(promises);
}

/*Calling fn, getting results in then since Promise.all() itself is promise*/
displayCountTravels()
  .then((results) => {
    /*using map to get lengths of the documents returned and put it in qweek*/
    var qweek = results.map((e) => e.length);
    console.log(qweek);
    console.log(qweek[1]);
  })
  .catch((e) => {
    console.error(e);
  });

Upvotes: 1

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