Using findOne in mongodb to get element with max id

Question

I am trying to retrieve one element from a mongo collection, the one with the greatest _id field. I know this can be done by querying:

db.collection.find().s         


        

Answer 1

You can get max _id using aggregation of mongodb. Find and sort may overkill's.

db.myCollection.aggregate({
    $group: {
        _id: '',
        last: {
            $max: "$_id"
        }
    }
});

Answer 2

with PHP driver (mongodb)
using findOne()

$filter=[];
$options = ['sort' => ['_id' => -1]]; // -1 is for DESC
$result = $collection->findOne(filter, $options);
$maxAge = $result['age']

Answer 3

import pymongo

tonystark = pymongo.MongoClient("mongodb://localhost:27017/")

mydb = tonystark["tonystark_db"]
savings = mydb["customers"]

x = savings.find().sort("_id")
for s in x:
    print(s)

Answer 4

$maxId="";

$Cursor =$collection->find();

foreach($cursor as $document) { 
    $maxid =max($arr=(array($document['id'])));
}

print_r($maxid+1);

Answer 5

You should use find, like you already are, and not aggregation which will be slower since it needs to scan all the values of _id fields to figure out the max.

As comments pointed out there is no difference between using find() and findOne() - functionally or elegance-wise. In fact, findOne in the shell (and in the drivers which implement it) is defined in terms of find (with limit -1 and with pretty print in the shell).

If you really want to do the equivalent of

db.collection.find().sort({_id:-1}).limit(1).pretty()

as findOne you can do it with this syntax:

db.collection.findOne({$query:{},$orderby:{_id:-1}})