How to sort by defined value in mongodb - spring

Question

In my project I'm using SpringBoot 1.4.0-M2 and MongoDb 3.0
I want to implement sorting.
I have collection of documents:
{
"name":"Tim",
"age":19
}
,{
"name":"Jim",
"age":20
}
,{
"name":"Will",
"age":21
}
,{
"name":"Ed",
"age":22
}
I random number from 19 - 22, lets pretend that a number I have got is 21.
Now I want to order by "age" property using my the 21 value.
I want to have order like this 21,22,19,20 so Will,Ed,Tim,Jim
Do You know how to sort in this way?
Best regards!

Answer 1

I don't know of a way to achieve this "wrapping sort" concept in a single query, but you could do it in two queries. The examples below load everything into memory, so you would just need to change things around a bit if you only wanted to operate on one at a time in the case of very large collections.
If you are using the native MongoDB Java driver 3.x:
int age = 21;
DBObject ageSort = new BasicDBObject("age", 1)
List<Document> results = new ArrayList<>();
mongoCollection.find(new BasicDBObject("age", new BasicDBObject("$gte", age)))
.sort(ageSort)
.forEach((Block<Document>) results::add);
mongoCollection.find(new BasicDBObject("age", new BasicDBObject("$lt", age)))
.sort(ageSort)
.forEach((Block<Document>) results::add);
If you are using Spring Data MongoDB:
int age = 21;
Sort ageSort = new Sort(new Sort.Order(Sort.Direction.ASC, "age"))
List<Map> results = mongoTemplate.findAll(
new Query(Criteria.where("age").gte(age)).with(ageSort),
Map.class
);
results.addAll(mongoTemplate.findAll(
new Query(Criteria.where("age").lt(age)).with(ageSort),
Map.class
));