I have a query in Mongoose that finds a set of objects, and then returns these said objects together with the total number of them:

var itemsPerPage = 4, pageNum = 1;

    .skip(itemsPerPage * pageNum)
    .then(docs => {
        if (docs.length === 0)
            console.log("There are no results matching your query.");
                .then(number => {
                    console.log(JSON.stringify({number, docs}));

To achieve this I do:

  1. find query with params
  2. use limit and skip to obtain results from correct page
  3. if docs is not empty, start a count query
  4. return information


The problem here, since I need pagination in the query, is that I have to make 2 separate requests to the DB (find and count) which can potential be very heavy.

To optimize this, I would like to avoid the count query, but I don't know how.

Another issue is the nested promises anti pattern, but that is not the issue I would like to focus on right now.


I am open to any suggestions on improving this code you may have!


1 Answer 1



After reviewing many answers, I believe I have two options:

  1. Make two queries, like I am doing now,
  2. Use a pagination plugin for mongoose

My decision was the second one and so I am now using mongoose-paginate.

Hope it helps.

  • \$\begingroup\$ I am still using the 2 query method up to now. I wish there was an auto index for this kind of total count. \$\endgroup\$
    – kabuto178
    Commented Sep 21, 2017 at 0:48
  • \$\begingroup\$ This lib mongoose-paginate does the same 2 queries: run the query first and again with count. promises = { docs: docsQuery.exec(), count: this.count(query).exec() }; \$\endgroup\$
    – jpenna
    Commented May 18, 2019 at 13:42
  • \$\begingroup\$ And count is deprecated, by the way. Use countDocuments or estimateCountDocuments if it is a big set. Mongoose count \$\endgroup\$
    – jpenna
    Commented May 19, 2019 at 4:03

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