How to join query in mongodb?

To have everything with just one query using the $lookup feature of the aggregation framework, try this :

db.User.aggregate(
    [
        // First step is to extract the "friends" field to work with the values
        {
            $unwind: "$friends"
        },
        // Lookup all the linked friends from the User collection
        {
            $lookup:
            {
                from: "User",
                localField: "friends",
                foreignField: "_id",
                as: "friendsData"
            }
        },
        // Sort the results by age
        {
            $sort: { 'friendsData.age': 1 }
        },
        // Get the results into a single array
        {
            $unwind: "$friendsData"
        },
        // Group the friends by user id
        {
            $group:
            {
                _id: "$_id",
                friends: { $push: "$friends" },
                friendsData: { $push: "$friendsData" }
            }
        }
    ]
)

Let's say the content of your User collection is the following:

{
    "_id" : ObjectId("573b09e6322304d5e7c6256e"),
    "name" : "John",
    "age" : 30,
    "friends" : [
        "userId1",
        "userId2",
        "userId3"
    ]
}
{ "_id" : "userId1", "name" : "Derek", "age" : 34 }
{ "_id" : "userId2", "name" : "Homer", "age" : 44 }
{ "_id" : "userId3", "name" : "Bobby", "age" : 12 }

The result of the query will be:

{
    "_id" : ObjectId("573b09e6322304d5e7c6256e"),
    "friends" : [
        "userId3",
        "userId1",
        "userId2"
    ],
    "friendsData" : [
        {
            "_id" : "userId3",
            "name" : "Bobby",
            "age" : 12
        },
        {
            "_id" : "userId1",
            "name" : "Derek",
            "age" : 34
        },
        {
            "_id" : "userId2",
            "name" : "Homer",
            "age" : 44
        }
    ]
}

Edit: this answer only applies to versions of MongoDb prior to v3.2.

You can't do what you want in just one query. You would have to first retrieve the list of friend user ids, then pass those ids to the second query to retrieve the documents and sort them by age.

var user = db.user.findOne({"id" : "001"}, {"friends": 1})
db.user.find( {"id" : {$in : user.friends }}).sort("age" : 1);