How to get the last N records in mongodb?

Question

I can't find anywhere it has been documented this. By default, the find() operation will get the records from beginning. How can I get the last N records in mongodb?

Edit: also I want the returned result ordered from less recent to most recent, not the reverse.

Answer 1

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();

Answer 2

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)

Answer 3

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);

Answer 4

Last function should be sort, not limit.

Example:

db.testcollection.find().limit(3).sort({timestamp:-1}); 

Answer 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.

Answer 6

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.