MongoDB: How to find nth highest salary from collection - mongodb

Question

I need to find the nth highest salary in a mongodb from Employees collection.
also would be really helpful if someone could gimme an idea of applying joins in mongodb.

Answer 1

This should work
db.Employees.find({}).sort({"Emp salary":-1}).limit(1) //for first highest salary
db.Employees.find({}).sort({"Emp salary":-1}).skip(1).limit(1) // for second highest salary
Similarly you can do db.Employees.find({}).sort({"Emp salary":-1}).skip(nthVarible - 1).limit(1).

Answer 2

Try this out:
db.salary.find({}).sort({s:-1}).skip(1).limit(1);
For your second requirement - MongoDB is noSQL DB, not a transnational DB. It does not support joins.

Answer 3

I found a two-step process to do it. it will work in the scenario if there are multiple records with salary same as highest.
Records I have
{ "_id" : ObjectId("5cc04b02536dc2e493697b4e"), "name" : "Ankit" }
{ "_id" : ObjectId("5cc0504a536dc2e493697b50"), "name" : "Ankit", "salary" : 1000, "email" : "a#b.com", "joining_date" : ISODate("2019-04-24T12:02:18.528Z") }
{ "_id" : ObjectId("5cc0504a536dc2e493697b51"), "name" : "Priya", "salary" : 1300, "email" : "p#b.com", "joining_date" : ISODate("2019-04-24T12:02:18.528Z") }
{ "_id" : ObjectId("5cc0504a536dc2e493697b52"), "name" : "Raj", "salary" : 1200, "email" : "rj#b.com", "joining_date" : ISODate("2019-04-24T12:02:18.528Z") }
{ "_id" : ObjectId("5cc0504a536dc2e493697b53"), "name" : "Vishu", "salary" : 1500, "email" : "v#b.com", "joining_date" : ISODate("2019-04-24T12:02:18.528Z") }
{ "_id" : ObjectId("5cc0504a536dc2e493697b54"), "name" : "Rahul", "salary" : 2000, "email" : "ra#b.com", "joining_date" : ISODate("2019-04-24T12:02:18.528Z") }
{ "_id" : ObjectId("5cc08b5d536dc2e493697b57"), "name" : "Tushar", "salary" : 2000, "email" : "tu#b.com", "joining_date" : ISODate("2019-04-24T16:14:21.061Z") }
Find distinct salaries and store in a variable
sal = db.employee.distinct("salary").sort()
Output: [ 1000, 1200, 1300, 1500, 2000 ]
You can get the second highest salary from this array itself. Below query will give you the record with that salary
db.employee.find({salary:{$lt:sal[sal.length-1]}}).sort({"salary":-1}).limit(1)
Output:
{ "_id" : ObjectId("5cc0504a536dc2e493697b53"), "name" : "Vishu", "salary" : 1500, "email" : "v#b.com", "joining_date" : ISODate("2019-04-24T12:02:18.528Z") }

Answer 4

I see this question has been asked in many technical interviews.
OBJ = client.my_db.employee_table
OBJ.find({}).sort('salary', -1).limit(1)
1 in the index 1 implies ascending order
-1 in the index 1 implies descending order.
As we wanna find the highest salary from the table, we must mention -1.
To find the nth highest salary in the table.
OBJ.find({}).sort('salary',-1).skip(n-1).limit(1)
To eliminate rows, we use OFFSET in Mysql/SQL Likewise we must use skip() in MongoDB

Answer 5

For first, second maximum salary if having multiple records in Mongodb:
METHOD I : (If NO multiple records exists)
db.details.find({}).sort({"salary":-1}).limit(1) ==> First Highest Salary
db.details.find({}).sort({"salary":-1}).skip(1).limit(1) ==> Second Highest Salary
METHOD II : (If MULTIPLES records exists)
Second Maximum Salary :
sal = db.details.distinct("salary").sort() ==> sal = [1000, 1400, 1500, 1700, 2000]
db.details.find({salary:{$lt:sal[sal.length-1]}}).sort({"salary":-1}).limit(1)

Answer 6

db.sales.aggregate({$group:{_id:'$salary'}},{$sort:{salary:-1}},{$skip: 1},{$limit:1})
In $skip value you can use n number to skip n number of rows.