how to find a single field in mongoDB instead of entire document - node.js

Question

I want to achieve the result as obtained by
SELECT AGE FROM COLL WHERE NAME="AYUSH";
I took the following approach
var MongoClient = require('mongodb').MongoClient;
var url = "mongodb://localhost:27017/";
MongoClient.connect(url, function(err, db) {
if (err) throw err;
var dbo = db.db("new");
//var query = { name:'ayush' };
//var age = {age : 1, _id:0};
dbo.collection("coll").find(
{ name:'ayush' },
{ age : 1, _id:0}).toArray(function(err, result) {
if (err) throw err;
console.log(result);
db.close();
});
});
the result that i am getting is
[ { _id: 5a818b71d2029813505d736a,
name: 'ayush',
age: '22',
sex: 'm' } ]

Answer 1

From MongoDB documentation : Project Fields to Return from Query
Return the Specified Fields and the _id Field Only
A projection can explicitly include several fields by setting the
<field> to 1 in the projection document. The following operation
returns all documents that match the query. In the result set, only
the item, status and, by default, the _id fields return in the
matching documents.
db.inventory.find( { status: "A" }, { item: 1, status: 1 } )
The operation corresponds to the following SQL statement:
SELECT _id, item, status from inventory WHERE status = "A"
In your case, if you only want the field age, you have to suppress the other fields _id, name and sex as following :
dbo.collection("coll").find({ name:'ayush' },{age:1, _id:0, name:0, sex:0})...

Answer 2

The easiest way to do this would be to convert the result in to an array and then send that array as the response.
The code would look like:
dbo.collection("coll").find(
{ name:'ayush' },
{ age : 1, _id:0}).toArray(function(err, result) {
if (err) throw err;
var array = [];
array.push(result[0].age);
res.send(array);});
Moreover don't use mongoClient,use mongoose instead.its easier and better