How to paginate with Mongoose in Node.js? - mongodb

Question

I am writing a webapp with Node.js and mongoose. How can I paginate the results I get from a .find() call? I would like a functionality comparable to "LIMIT 50,100" in SQL.

Answer 1

I'm am very disappointed by the accepted answers in this question. This will not scale. If you read the fine print on cursor.skip( ):
The cursor.skip() method is often expensive because it requires the server to walk from the beginning of the collection or index to get the offset or skip position before beginning to return result. As offset (e.g. pageNumber above) increases, cursor.skip() will become slower and more CPU intensive. With larger collections, cursor.skip() may become IO bound.
To achieve pagination in a scaleable way combine a limit( ) along with at least one filter criterion, a createdOn date suits many purposes.
MyModel.find( { createdOn: { $lte: request.createdOnBefore } } )
.limit( 10 )
.sort( '-createdOn' )

Answer 2

After taking a closer look at the Mongoose API with the information provided by Rodolphe, I figured out this solution:
MyModel.find(query, fields, { skip: 10, limit: 5 }, function(err, results) { ... });

Answer 3

Pagination using mongoose, express and jade - Here's a link to my blog with more detail
var perPage = 10
, page = Math.max(0, req.params.page)
Event.find()
.select('name')
.limit(perPage)
.skip(perPage * page)
.sort({
name: 'asc'
})
.exec(function(err, events) {
Event.count().exec(function(err, count) {
res.render('events', {
events: events,
page: page,
pages: count / perPage
})
})
})

Answer 4

You can chain just like that:
var query = Model.find().sort('mykey', 1).skip(2).limit(5)
Execute the query using exec
query.exec(callback);

Answer 5

In this case, you can add the query page and/ or limit to your URL as a query string.
For example:
?page=0&limit=25 // this would be added onto your URL: http:localhost:5000?page=0&limit=25
Since it would be a String we need to convert it to a Number for our calculations. Let's do it using the parseInt method and let's also provide some default values.
const pageOptions = {
page: parseInt(req.query.page, 10) || 0,
limit: parseInt(req.query.limit, 10) || 10
}
sexyModel.find()
.skip(pageOptions.page * pageOptions.limit)
.limit(pageOptions.limit)
.exec(function (err, doc) {
if(err) { res.status(500).json(err); return; };
res.status(200).json(doc);
});
BTW
Pagination starts with 0

Answer 6

You can use a little package called Mongoose Paginate that makes it easier.
$ npm install mongoose-paginate
After in your routes or controller, just add :
/**
* querying for `all` {} items in `MyModel`
* paginating by second page, 10 items per page (10 results, page 2)
**/
MyModel.paginate({}, 2, 10, function(error, pageCount, paginatedResults) {
if (error) {
console.error(error);
} else {
console.log('Pages:', pageCount);
console.log(paginatedResults);
}
}

Answer 7

Query:
search = productName
Params:
page = 1
// Pagination
router.get("/search/:page", (req, res, next) => {
const resultsPerPage = 5;
let page = req.params.page >= 1 ? req.params.page : 1;
const query = req.query.search;
page = page - 1
Product.find({ name: query })
.select("name")
.sort({ name: "asc" })
.limit(resultsPerPage)
.skip(resultsPerPage * page)
.then((results) => {
return res.status(200).send(results);
})
.catch((err) => {
return res.status(500).send(err);
});
});

Answer 8

This is a example you can try this,
var _pageNumber = 2,
_pageSize = 50;
Student.count({},function(err,count){
Student.find({}, null, {
sort: {
Name: 1
}
}).skip(_pageNumber > 0 ? ((_pageNumber - 1) * _pageSize) : 0).limit(_pageSize).exec(function(err, docs) {
if (err)
res.json(err);
else
res.json({
"TotalCount": count,
"_Array": docs
});
});
});

Answer 9

Try using mongoose function for pagination. Limit is the number of records per page and number of the page.
var limit = parseInt(body.limit);
var skip = (parseInt(body.page)-1) * parseInt(limit);
db.Rankings.find({})
.sort('-id')
.limit(limit)
.skip(skip)
.exec(function(err,wins){
});

Answer 10

This is what I done it on code
var paginate = 20;
var page = pageNumber;
MySchema.find({}).sort('mykey', 1).skip((pageNumber-1)*paginate).limit(paginate)
.exec(function(err, result) {
// Write some stuff here
});
That is how I done it.

Answer 11

Simple and powerful pagination solution
async getNextDocs(no_of_docs_required: number = 5, last_doc_id?: string) {
let docs
if (!last_doc_id) {
// get first 5 docs
docs = await MySchema.find().sort({ _id: -1 }).limit(no_of_docs_required)
}
else {
// get next 5 docs according to that last document id
docs = await MySchema.find({_id: {$lt: last_doc_id}})
.sort({ _id: -1 }).limit(no_of_docs_required)
}
return docs
}
last_doc_id: the last document id that you get
no_of_docs_required: the number of docs that you want to fetch i.e. 5, 10, 50 etc.
If you don't provide the last_doc_id to the method, you'll get i.e. 5 latest docs
If you've provided the last_doc_id then you'll get the next i.e. 5 documents.

Answer 12

There are some good answers giving the solution that uses skip() & limit(), however, in some scenarios, we also need documents count to generate pagination. Here's what we do in our projects:
const PaginatePlugin = (schema, options) => {
options = options || {}
schema.query.paginate = async function(params) {
const pagination = {
limit: options.limit || 10,
page: 1,
count: 0
}
pagination.limit = parseInt(params.limit) || pagination.limit
const page = parseInt(params.page)
pagination.page = page > 0 ? page : pagination.page
const offset = (pagination.page - 1) * pagination.limit
const [data, count] = await Promise.all([
this.limit(pagination.limit).skip(offset),
this.model.countDocuments(this.getQuery())
]);
pagination.count = count;
return { data, pagination }
}
}
mySchema.plugin(PaginatePlugin, { limit: DEFAULT_LIMIT })
// using async/await
const { data, pagination } = await MyModel.find(...)
.populate(...)
.sort(...)
.paginate({ page: 1, limit: 10 })
// or using Promise
MyModel.find(...).paginate(req.query)
.then(({ data, pagination }) => {
})
.catch(err => {
})

Answer 13

Here is a version that I attach to all my models. It depends on underscore for convenience and async for performance. The opts allows for field selection and sorting using the mongoose syntax.
var _ = require('underscore');
var async = require('async');
function findPaginated(filter, opts, cb) {
var defaults = {skip : 0, limit : 10};
opts = _.extend({}, defaults, opts);
filter = _.extend({}, filter);
var cntQry = this.find(filter);
var qry = this.find(filter);
if (opts.sort) {
qry = qry.sort(opts.sort);
}
if (opts.fields) {
qry = qry.select(opts.fields);
}
qry = qry.limit(opts.limit).skip(opts.skip);
async.parallel(
[
function (cb) {
cntQry.count(cb);
},
function (cb) {
qry.exec(cb);
}
],
function (err, results) {
if (err) return cb(err);
var count = 0, ret = [];
_.each(results, function (r) {
if (typeof(r) == 'number') {
count = r;
} else if (typeof(r) != 'number') {
ret = r;
}
});
cb(null, {totalCount : count, results : ret});
}
);
return qry;
}
Attach it to your model schema.
MySchema.statics.findPaginated = findPaginated;

Answer 14

Above answer's holds good.
Just an add-on for anyone who is into async-await rather than
promise !!
const findAllFoo = async (req, resp, next) => {
const pageSize = 10;
const currentPage = 1;
try {
const foos = await FooModel.find() // find all documents
.skip(pageSize * (currentPage - 1)) // we will not retrieve all records, but will skip first 'n' records
.limit(pageSize); // will limit/restrict the number of records to display
const numberOfFoos = await FooModel.countDocuments(); // count the number of records for that model
resp.setHeader('max-records', numberOfFoos);
resp.status(200).json(foos);
} catch (err) {
resp.status(500).json({
message: err
});
}
};

Answer 15

you can use the following line of code as well
per_page = parseInt(req.query.per_page) || 10
page_no = parseInt(req.query.page_no) || 1
var pagination = {
limit: per_page ,
skip:per_page * (page_no - 1)
}
users = await User.find({<CONDITION>}).limit(pagination.limit).skip(pagination.skip).exec()
this code will work in latest version of mongo

Answer 16

A solid approach to implement this would be to pass the values from the frontend using a query string. Let's say we want to get page #2 and also limit the output to 25 results.
The query string would look like this: ?page=2&limit=25 // this would be added onto your URL: http:localhost:5000?page=2&limit=25
Let's see the code:
// We would receive the values with req.query.<<valueName>> => e.g. req.query.page
// Since it would be a String we need to convert it to a Number in order to do our
// necessary calculations. Let's do it using the parseInt() method and let's also provide some default values:
const page = parseInt(req.query.page, 10) || 1; // getting the 'page' value
const limit = parseInt(req.query.limit, 10) || 25; // getting the 'limit' value
const startIndex = (page - 1) * limit; // this is how we would calculate the start index aka the SKIP value
const endIndex = page * limit; // this is how we would calculate the end index
// We also need the 'total' and we can get it easily using the Mongoose built-in **countDocuments** method
const total = await <<modelName>>.countDocuments();
// skip() will return a certain number of results after a certain number of documents.
// limit() is used to specify the maximum number of results to be returned.
// Let's assume that both are set (if that's not the case, the default value will be used for)
query = query.skip(startIndex).limit(limit);
// Executing the query
const results = await query;
// Pagination result
// Let's now prepare an object for the frontend
const pagination = {};
// If the endIndex is smaller than the total number of documents, we have a next page
if (endIndex < total) {
pagination.next = {
page: page + 1,
limit
};
}
// If the startIndex is greater than 0, we have a previous page
if (startIndex > 0) {
pagination.prev = {
page: page - 1,
limit
};
}
// Implementing some final touches and making a successful response (Express.js)
const advancedResults = {
success: true,
count: results.length,
pagination,
data: results
}
// That's it. All we have to do now is send the `results` to the frontend.
res.status(200).json(advancedResults);
I would suggest implementing this logic into middleware so you can be able to use it for various routes/ controllers.

Answer 17

You can do using mongoose-paginate-v2. For more info click here
const mongoose = require('mongoose');
const mongoosePaginate = require('mongoose-paginate-v2');
const mySchema = new mongoose.Schema({
// your schema code
});
mySchema.plugin(mongoosePaginate);
const myModel = mongoose.model('SampleModel', mySchema);
myModel.paginate().then({}) // Usage

Answer 18

I have found a very efficient way and implemented it myself, I think this way is the best for the following reasons:
It does not use skip, which time complexity doesn't scale well;
It uses IDs to query the document. Ids are indexed by default in MongoDB, making them very fast to query;
It uses lean queries, these are known to be VERY performative, as they remove a lot of "magic" from Mongoose and returns a document that comes kind of "raw" from MongoDB;
It doesn't depend on any third party packages that might contain vulnerabilities or have vulnerable dependencies.
The only caveat to this is that some methods of Mongoose, such as .save() will not work well with lean queries, such methods are listed in this awesome blog post, I really recommend this series, because it considers a lot of aspects, such as type security (which prevents critical errors) and PUT/ PATCH.
I will provide some context, this is a Pokémon repository, the pagination works as the following: The API receives unsafeId from the req.body object of Express, we need to convert this to string in order to prevent NoSQL injections (it could be an object with evil filters), this unsafeId can be an empty string or the ID of the last item of the previous page, it goes like this:
/**
* #description GET All with pagination, will return 200 in success
* and receives the last ID of the previous page or undefined for the first page
* Note: You should take care, read and consider about Off-By-One error
* #param {string|undefined|unknown} unsafeId - An entire page that comes after this ID will be returned
*/
async readPages(unsafeId) {
try {
const id = String(unsafeId || '');
let criteria;
if (id) {
criteria = {_id: {$gt: id}};
} // else criteria is undefined
// This query looks a bit redundant on `lean`, I just really wanted to make sure it is lean
const pokemon = await PokemonSchema.find(
criteria || {},
).setOptions({lean: true}).limit(15).lean();
// This would throw on an empty page
// if (pokemon.length < 1) {
// throw new PokemonNotFound();
// }
return pokemon;
} catch (error) {
// In this implementation, any error that is not defined by us
// will not return on the API to prevent information disclosure.
// our errors have this property, that indicate
// that no sensitive information is contained within this object
if (error.returnErrorResponse) {
throw error;
} // else
console.error(error.message);
throw new InternalServerError();
}
}
Now, to consume this and avoid Off-By-One errors in the frontend, you do it like the following, considering that pokemons is the Array of Pokémons documents that are returned from the API:
// Page zero
const pokemons = await fetchWithPagination({'page': undefined});
// Page one
// You can also use a fixed number of pages instead of `pokemons.length`
// But `pokemon.length` is more reliable (and a bit slower)
// You will have trouble with the last page if you use it with a constant
// predefined number
const id = pokemons[pokemons.length - 1]._id;
if (!id) {
throw new Error('Last element from page zero has no ID');
} // else
const page2 = await fetchWithPagination({'page': id});
As a note here, Mongoose IDs are always sequential, this means that any newer ID will always be greater than the older one, that is the foundation of this answer.
This approach has been tested agaisnt Off-By-One errors, for instance, the last element of a page could be returned as the first element of the following one (duplicated), or an element that is between the last of the previous page and the first of the current page might disappear.
When you are done with all the pages and request a page after the last element (one that does not exist), the response will be an empty array with 200 (OK), which is awesome!

Answer 19

The easiest and more speedy way is, paginate with the objectId
Example;
Initial load condition
condition = {limit:12, type:""};
Take the first and last ObjectId from response data
Page next condition
condition = {limit:12, type:"next", firstId:"57762a4c875adce3c38c662d", lastId:"57762a4c875adce3c38c6615"};
Page next condition
condition = {limit:12, type:"next", firstId:"57762a4c875adce3c38c6645", lastId:"57762a4c875adce3c38c6675"};
In mongoose
var condition = {};
var sort = { _id: 1 };
if (req.body.type == "next") {
condition._id = { $gt: req.body.lastId };
} else if (req.body.type == "prev") {
sort = { _id: -1 };
condition._id = { $lt: req.body.firstId };
}
var query = Model.find(condition, {}, { sort: sort }).limit(req.body.limit);
query.exec(function(err, properties) {
return res.json({ "result": result);
});

Answer 20

The best approach (IMO) is to use skip and limit BUT within a limited collections or documents.
To make the query within limited documents, we can use specific index like index on a DATE type field. See that below
let page = ctx.request.body.page || 1
let size = ctx.request.body.size || 10
let DATE_FROM = ctx.request.body.date_from
let DATE_TO = ctx.request.body.date_to
var start = (parseInt(page) - 1) * parseInt(size)
let result = await Model.find({ created_at: { $lte: DATE_FROM, $gte: DATE_TO } })
.sort({ _id: -1 })
.select('<fields>')
.skip( start )
.limit( size )
.exec(callback)

Answer 21

Most easiest plugin for pagination.
https://www.npmjs.com/package/mongoose-paginate-v2
Add plugin to a schema and then use model paginate method:
var mongoose = require('mongoose');
var mongoosePaginate = require('mongoose-paginate-v2');
var mySchema = new mongoose.Schema({
/* your schema definition */
});
mySchema.plugin(mongoosePaginate);
var myModel = mongoose.model('SampleModel', mySchema);
myModel.paginate().then({}) // Usage

Answer 22

let page,limit,skip,lastPage, query;
page = req.params.page *1 || 1; //This is the page,fetch from the server
limit = req.params.limit * 1 || 1; // This is the limit ,it also fetch from the server
skip = (page - 1) * limit; // Number of skip document
lastPage = page * limit; //last index
counts = await userModel.countDocuments() //Number of document in the collection
query = query.skip(skip).limit(limit) //current page
const paginate = {}
//For previous page
if(skip > 0) {
paginate.prev = {
page: page - 1,
limit: limit
}
//For next page
if(lastPage < counts) {
paginate.next = {
page: page + 1,
limit: limit
}
results = await query //Here is the final results of the query.

Answer 23

const page = req.query.page * 1 || 1;
const limit = req.query.limit * 1 || 1000;
const skip = (page - 1) * limit;
query = query.skip(skip).limit(limit);

Answer 24

This is example function for getting the result of skills model with pagination and limit options
export function get_skills(req, res){
console.log('get_skills');
var page = req.body.page; // 1 or 2
var size = req.body.size; // 5 or 10 per page
var query = {};
if(page < 0 || page === 0)
{
result = {'status': 401,'message':'invalid page number,should start with 1'};
return res.json(result);
}
query.skip = size * (page - 1)
query.limit = size
Skills.count({},function(err1,tot_count){ //to get the total count of skills
if(err1)
{
res.json({
status: 401,
message:'something went wrong!',
err: err,
})
}
else
{
Skills.find({},{},query).sort({'name':1}).exec(function(err,skill_doc){
if(!err)
{
res.json({
status: 200,
message:'Skills list',
data: data,
tot_count: tot_count,
})
}
else
{
res.json({
status: 401,
message: 'something went wrong',
err: err
})
}
}) //Skills.find end
}
});//Skills.count end
}

Answer 25

Using ts-mongoose-pagination
const trainers = await Trainer.paginate(
{ user: req.userId },
{
perPage: 3,
page: 1,
select: '-password, -createdAt -updatedAt -__v',
sort: { createdAt: -1 },
}
)
return res.status(200).json(trainers)

Answer 26

Below Code Is Working Fine For Me.
You can add finding filters also and user same in countDocs query to get accurate results.
export const yourController = async (req, res) => {
const { body } = req;
var perPage = body.limit,
var page = Math.max(0, body.page);
yourModel
.find() // You Can Add Your Filters inside
.limit(perPage)
.skip(perPage * (page - 1))
.exec(function (err, dbRes) {
yourModel.count().exec(function (err, count) { // You Can Add Your Filters inside
res.send(
JSON.stringify({
Articles: dbRes,
page: page,
pages: count / perPage,
})
);
});
});
};

Answer 27

You can write query like this.
mySchema.find().skip((page-1)*per_page).limit(per_page).exec(function(err, articles) {
if (err) {
return res.status(400).send({
message: err
});
} else {
res.json(articles);
}
});
page : page number coming from client as request parameters.
per_page : no of results shown per page
If you are using MEAN stack following blog post provides much of the information to create pagination in front end using angular-UI bootstrap and using mongoose skip and limit methods in the backend.
see : https://techpituwa.wordpress.com/2015/06/06/mean-js-pagination-with-angular-ui-bootstrap/

Answer 28

You can either use skip() and limit(), but it's very inefficient. A better solution would be a sort on indexed field plus limit().
We at Wunderflats have published a small lib here: https://github.com/wunderflats/goosepage
It uses the first way.

Answer 29

If you are using mongoose as a source for a restful api have a look at
'restify-mongoose' and its queries. It has exactly this functionality built in.
Any query on a collection provides headers that are helpful here
test-01:~$ curl -s -D - localhost:3330/data?sort=-created -o /dev/null
HTTP/1.1 200 OK
link: </data?sort=-created&p=0>; rel="first", </data?sort=-created&p=1>; rel="next", </data?sort=-created&p=134715>; rel="last"
.....
Response-Time: 37
So basically you get a generic server with a relatively linear load time for queries to collections. That is awesome and something to look at if you want to go into a own implementation.

Answer 30

app.get("/:page",(req,res)=>{
post.find({}).then((data)=>{
let per_page = 5;
let num_page = Number(req.params.page);
let max_pages = Math.ceil(data.length/per_page);
if(num_page == 0 || num_page > max_pages){
res.render('404');
}else{
let starting = per_page*(num_page-1)
let ending = per_page+starting
res.render('posts', {posts:data.slice(starting,ending), pages: max_pages, current_page: num_page});
}
});
});