'How to make pagination with mongoose
I want to make a pagination feature in my Collection. How can find a documents with 'start' and 'limit' positions and get total document number in a single query?
Solution 1:[1]
You can use the plugin Mongoose Paginate:
$ npm install mongoose-paginate
After in your routes or controller, just add :
Model.paginate({}, { page: 3, limit: 10 }, function(err, result) {
// result.docs
// result.total
// result.limit - 10
// result.page - 3
// result.pages
});
Solution 2:[2]
If you plan to have a lot of pages, you should not use skip/limit, but rather calculate ranges.
See Scott's answer for a similar question: MongoDB - paging
Solution 3:[3]
UPDATE :
Using skip and limit is not good for pagination. Here is the discussion over it.
@Wes Freeman, Gave a good answer. I read the linked pose, you should use range query. n = 0; i = n+10; db.students.find({ "id" : { $gt: n, $lt: (n+i)} } );
OLD ANSWER (don't use this)
use something like
n = db.students.find().skip(n).limit(10);
//pass n dynamically, so for page 1 it should be 0 , page 2 it should be 10 etc
more documentation at http://www.mongodb.org/display/DOCS/Advanced+Queries
Solution 4:[4]
user.find({_id:{$nin:friends_id},_id:{$ne:userId}},function(err,user_details){
if (err)
res.send(err);
response ={
statusCode:200,
status:true,
values:user_details
}
res.json(response);
}).skip(10).limit(1);
Solution 5:[5]
I am using this function,
You can check if prev and next data is available or not.
async (req, res) => {
let { page = 1, size = 10 } = req.query
page = parseInt(page)
size = parseInt(size)
const query = {}
const totalData = await Model.find().estimatedDocumentCount()
const data = await Model.find(query).skip((page - 1) * size).limit(size).exec()
const pageNumber = Math.ceil(totalData / size)
const results = {
currentPage: page,
prevPage: page <= 1 ? null : page - 1,
nextPage: page >= pageNumber ? null : page + 1,
data
}
res.json(results) }
To know more estimatedDocumentCount
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | Community |
| Solution 3 | Community |
| Solution 4 | RBJT |
| Solution 5 | Najmus Sakib |
