How to get the last N records in mongodb?

mongodb record

If I understand your question, you need to sort in ascending order.

Assuming you have some id or date field called "x" you would do ...

.sort()

db.foo.find().sort({x:1});

The 1 will sort ascending (oldest to newest) and -1 will sort descending (newest to oldest.)

If you use the auto created _id field it has a date embedded in it ... so you can use that to order by ...

db.foo.find().sort({_id:1});

That will return back all your documents sorted from oldest to newest.

You can also use a Natural Order mentioned above ...

db.foo.find().sort({$natural:1});

Again, using 1 or -1 depending on the order you want.

Lastly, it's good practice to add a limit when doing this sort of wide open query so you could do either ...

db.foo.find().sort({_id:1}).limit(50);

db.foo.find().sort({$natural:1}).limit(50);

mongodb record

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)

mongodb record

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.