在MongoDB中查找最新对象

I have data in MongoDB in the form of :

 {
            "_id" : 154,
            "record_id" : "001280000033x54AAA",
            "jsonData" : "",
            "user_id" : 1,
            "userName" : "abc@xyz.com",
            "backup_no" : 1
 }


 {
            "_id" : 155,
            "record_id" : "001280000033x54AAA",
            "jsonData" : "",
            "user_id" : 1,
            "userName" : "abc@xyz.com",
            "backup_no" : 2
 }
  ...

I want to retrieve data based on 'user_id','userName', but if a record with same record_id exists in lower 'backup_no' then i need to choose record with highest backup_no.

I have tried to aggregate record_id's and then query but i am unable to find a solution.

Thanks

Using mongo aggregation you can get your results check below query :

db.collectionName.aggregate({
  "$sort": {
    "backup_no": -1 //first sort by backup_no 
  }
}, {
  "$group": {
    "_id": "$record_id", // group by record_id so here only get distinct record_id
    "userId": {
      "$first": "$user_id"
    },
    "userName": {
      "$first": "$userName"
    },
    "backup_no": {
      "$first": "$backup_no"
    }
  }
}, {
  "$project": {
    "_id": 0,
    "record_id": "$_id",
    "user_id": "$user_id",
    "userName": "$userName",
    "backup_no": "$backup_no"
  }
})

db.collection.aggregate([{
  $group: {
    _id: "$record_id",
    user_id: {
      $last: "$user_id"
    },
    userName: {
      $last: "$userName"
    },
    backup_no: {
      $max: "$backup_no"
    }
  }
}])

or

You could simply sort descending by backup_no and then ignore all but the first record for each distinct record_id?
MongoDB $orderby

db.users.find( { } ).sort( { backup_no: -1 } );