Get total count along with Mongoose Query skip & limt - node.js

Question

I have a json data which contains many objects. I want to limit the data for pagination and I need the total items count. Please help.
Model.find().skip((pageNumber-1)*limit).limit(limit).exec()
I want the count and skipped data in response.

Answer 1

You can use async library for running 2 queries at once. In your case you can run one query to get the number of documents and another for pagination.
Example with 'User' model:
var async = require('async');
var User = require('./models/user');
var countQuery = function(callback){
User.count({}, function(err, count){
if(err){ callback(err, null) }
else{
callback(null, count);
}
}
};
var retrieveQuery = function(callback){
User.find({}).skip((page-1)*PAGE_LIMIT)
.limit(PAGE_LIMIT)
.exec(function(err, doc){
if(err){ callback(err, null) }
else{
callback(null, doc);
}
}
};
async.parallel([countQuery, retrieveQuery], function(err, results){
//err contains the array of error of all the functions
//results contains an array of all the results
//results[0] will contain value of doc.length from countQuery function
//results[1] will contain doc of retrieveQuery function
//You can send the results as
res.json({users: results[1], pageLimit: PAGE_LIMIT, page: page, totalCount: results[0]});
});
async allows you to run a number of queries in parallel depending on the hardware you are using. This would be faster than using 2 independent queries to get count and get the required documents.
Hope this helps.

Answer 2

I have solved it with $facet and aggregate the following way in mongoose v3+:
const [{ paginatedResult, [{ totalCount }] }] = await Model.aggregate([{
$facet: {
paginatedResult: [
{ $match: query },
{ $skip: skip },
{ $limit: limit }
],
totalCount: [
{ $match: query },
{ $count: 'totalCount' }
]
}
}])
where the totalCount refers the total number of records matching the search query while the paginatedResult is only the paginated slice of them.

Answer 3

The problem with these solutions is that for every request you are doing two queries. This becomes problematic when you have a complex data structure and large data set as performance becomes an issue. Consider instead creating a special function that listens for the /resource?count=true or /resource/count GET methods and returns only the count.

Answer 4

You need to perform 2 queries to achieve that. One to get results and another to get total items amount with .count().
For example code you can watch at on of "paginator" for mongoose mongoose-paginate.

Answer 5

To performe only one query, you may use the find() method associated with promises and array slices. A small example would be:
getPaginated(query, skip, limit){
return this.model.find(query)
.lean()
.then((value)=>{
if (value.length === 0) return {userMessage: 'Document not found'};
const count = value.length;
//skip===0 must be handled
const start = parseInt(limit)*parseInt(skip - 1);
const end = start + parseInt(reqQuery.pagesize);
//slicing the array
value = value.slice(start,end);
//could return it another way...
value.push( { 'querySize': count });
return value;
})
.catch((reason)=>{
//...handling code
});
}