Mongoose limit/offset and count query

There is a library that will do all of this for you, check out mongoose-paginate-v2

I suggest you to use 2 queries:

1. db.collection.count() will return total number of items. This value is stored somewhere in Mongo and it is not calculated.

2. db.collection.find().skip(20).limit(10) here I assume you could use a sort by some field, so do not forget to add an index on this field. This query will be fast too.

I think that you shouldn't query all items and than perform skip and take, cause later when you have big data you will have problems with data transferring and processing.

db.collection_name.aggregate([
    { '$match'    : { } },
    { '$sort'     : { '_id' : -1 } },
    { '$facet'    : {
        metadata: [ { $count: "total" } ],
        data: [ { $skip: 1 }, { $limit: 10 },{ '$project' : {"_id":0} } ] // add projection here wish you re-shape the docs
    } }
] )

Instead of using two queries to find the total count and skip the matched record.
$facet is the best and optimized way.

1. Match the record
   
2. Find total_count
   
3. skip the record
   
4. And also can reshape data according to our needs in the query.

Instead of using 2 separate queries, you can use aggregate() in a single query:

Aggregate "$facet" can be fetch more quickly, the Total Count and the Data with skip & limit

    db.collection.aggregate([

      //{$sort: {...}}

      //{$match:{...}}

      {$facet:{

        "stage1" : [ {"$group": {_id:null, count:{$sum:1}}} ],

        "stage2" : [ { "$skip": 0}, {"$limit": 2} ]
  
      }},
     
     {$unwind: "$stage1"},
  
      //output projection
     {$project:{
        count: "$stage1.count",
        data: "$stage2"
     }}

 ]);

output as follows:-

[{
     count: 50,
     data: [
        {...},
        {...}
      ]
 }]

Also, have a look at https://docs.mongodb.com/manual/reference/operator/aggregation/facet/