search in limited number of record MongoDB - mongodb

Question

I want to search in the first 1000 records of my document whose name is CityDB. I used the following code:
db.CityDB.find({'index.2':"London"}).limit(1000)
but it does not work, it return the first 1000 of finding, but I want to search just in the first 1000 records not all records. Could you please help me.
Thanks,
Amir

Answer 1

Note that there is no guarantee that your documents are returned in any particular order by a query as long as you don't sort explicitely. Documents in a new collection are usually returned in insertion order, but various things can cause that order to change unexpectedly, so don't rely on it. By the way: Auto-generated _id's start with a timestamp, so when you sort by _id, the objects are returned by creation-date.
Now about your actual question. When you first want to limit the documents and then perform a filter-operation on this limited set, you can use the aggregation pipeline. It allows you to use $limit-operator first and then use the $match-operator on the remaining documents.
db.CityDB.aggregate(
// { $sort: { _id: 1 } }, // <- uncomment when you want the first 1000 by creation-time
{ $limit: 1000 },
{ $match: { 'index.2':"London" } }
)

Answer 2

I can think of two ways to achieve this:
1) You have a global counter and every time you input data into your collection you add a field count = currentCounter and increase currentCounter by 1. When you need to select your first k elements, you find it this way
db.CityDB.find({
'index.2':"London",
count : {
'$gte' : currentCounter - k
}
})
This is not atomic and might give you sometimes more then k elements on a heavy loaded system (but it can support indexes).
Here is another approach which works nice in the shell:
2) Create your dummy data:
var k = 100;
for(var i = 1; i<k; i++){
db.a.insert({
_id : i,
z: Math.floor(1 + Math.random() * 10)
})
}
output = [];
And now find in the first k records where z == 3
k = 10;
db.a.find().sort({$natural : -1}).limit(k).forEach(function(el){
if (el.z == 3){
output.push(el)
}
})
as you see your output has correct elements:
output
I think it is pretty straight forward to modify my example for your needs.
P.S. also take a look in aggregation framework, there might be a way to achieve what you need with it.