How to get the last N records in mongodb?

The last N added records, from less recent to most recent, can be seen with this query:

db.collection.find().skip(db.collection.count() - N)

If you want them in the reverse order:

db.collection.find().sort({ $natural: -1 }).limit(N)

If you install Mongo-Hacker you can also use:

db.collection.find().reverse().limit(N)

If you get tired of writing these commands all the time you can create custom functions in your ~/.mongorc.js. E.g.

function last(N) {
    return db.collection.find().skip(db.collection.count() - N);
}

then from a mongo shell just type last(N)

In order to get last N records you can execute below query:

db.yourcollectionname.find({$query: {}, $orderby: {$natural : -1}}).limit(yournumber)

if you want only one last record:

db.yourcollectionname.findOne({$query: {}, $orderby: {$natural : -1}})

Note: In place of $natural you can use one of the columns from your collection.

Sorting, skipping and so on can be pretty slow depending on the size of your collection.

A better performance would be achieved if you have your collection indexed by some criteria; and then you could use min() cursor:

First, index your collection with db.collectionName.setIndex( yourIndex ) You can use ascending or descending order, which is cool, because you want always the "N last items"... so if you index by descending order it is the same as getting the "first N items".

Then you find the first item of your collection and use its index field values as the min criteria in a search like:

db.collectionName.find().min(minCriteria).hint(yourIndex).limit(N)

Here's the reference for min() cursor: https://docs.mongodb.com/manual/reference/method/cursor.min/

 db.collection.find().sort({$natural: -1 }).limit(5)

@bin-chen,

You can use an aggregation for the latest n entries of a subset of documents in a collection. Here's a simplified example without grouping (which you would be doing between stages 4 and 5 in this case).

This returns the latest 20 entries (based on a field called "timestamp"), sorted ascending. It then projects each documents _id, timestamp and whatever_field_you_want_to_show into the results.

var pipeline = [
        {
            "$match": { //stage 1: filter out a subset
                "first_field": "needs to have this value",
                "second_field": "needs to be this"
            }
        },
        {
            "$sort": { //stage 2: sort the remainder last-first
                "timestamp": -1
            }
        },
        {
            "$limit": 20 //stage 3: keep only 20 of the descending order subset
        },
        {
            "$sort": {
                "rt": 1 //stage 4: sort back to ascending order
            }
        },
        {
            "$project": { //stage 5: add any fields you want to show in your results
                "_id": 1,
                "timestamp" : 1,
                "whatever_field_you_want_to_show": 1
            }
        }
    ]

yourcollection.aggregate(pipeline, function resultCallBack(err, result) {
  // account for (err)
  // do something with (result)
}

so, result would look something like:

{ 
    "_id" : ObjectId("5ac5b878a1deg18asdafb060"),
    "timestamp" : "2018-04-05T05:47:37.045Z",
    "whatever_field_you_want_to_show" : -3.46000003814697
}
{ 
    "_id" : ObjectId("5ac5b878a1de1adsweafb05f"),
    "timestamp" : "2018-04-05T05:47:38.187Z",
    "whatever_field_you_want_to_show" : -4.13000011444092
}

Hope this helps.

You can try this method:

Get the total number of records in the collection with

db.dbcollection.count() 

Then use skip:

db.dbcollection.find().skip(db.dbcollection.count() - 1).pretty()

You can't "skip" based on the size of the collection, because it will not take the query conditions into account.

The correct solution is to sort from the desired end-point, limit the size of the result set, then adjust the order of the results if necessary.

Here is an example, based on real-world code.

var query = collection.find( { conditions } ).sort({$natural : -1}).limit(N);

query.exec(function(err, results) {
    if (err) { 
    }
    else if (results.length == 0) {
    }
    else {
        results.reverse(); // put the results into the desired order
        results.forEach(function(result) {
            // do something with each result
        });
    }
});

you can use sort() , limit() ,skip() to get last N record start from any skipped value

db.collections.find().sort(key:value).limit(int value).skip(some int value);

Look under Querying: Sorting and Natural Order, http://www.mongodb.org/display/DOCS/Sorting+and+Natural+Order as well as sort() under Cursor Methods http://www.mongodb.org/display/DOCS/Advanced+Queries

If you use MongoDB compass, you can use sort filed to filter,

Use .sort() and .limit() for that

Use Sort in ascending or descending order and then use limit

db.collection.find({}).sort({ any_field: -1 }).limit(last_n_records);

You may want to be using the find options : http://docs.meteor.com/api/collections.html#Mongo-Collection-find

db.collection.find({}, {sort: {createdAt: -1}, skip:2, limit: 18}).fetch();

db.collection.find().hint( { $natural : -1 } ).sort(field: 1/-1).limit(n)

according to mongoDB Documentation:

You can specify { $natural : 1 } to force the query to perform a forwards collection scan.

You can also specify { $natural : -1 } to force the query to perform a reverse collection scan.

use $slice operator to limit array elements

GeoLocation.find({},{name: 1, geolocation:{$slice: -5}})
    .then((result) => {
      res.json(result);
    })
    .catch((err) => {
      res.status(500).json({ success: false, msg: `Something went wrong. ${err}` });
});

where geolocation is array of data, from that we get last 5 record.