php - How to find the most current field that is embedded in an array for X user in mongodb

Question

I need to find a user with X _id and I need the most current weightTracker.date usually it will be the last one.

I've tried db.user.find({_id:userid}, {weightTracker:1}).sort({"weightTracker.date":-1}) With little success on ordering my results. But most important that always returns all the results and I am only interested in the most current one.

How Could I get the most recent object in weightTracker given the object's date in user X?

User

_id
weightTracker[{ manyproperties:menyproperties, date:date}]

Answer 1

您可能想尝试聚合方法（有关详细示例，您可以查看http://docs.mongodb.org/manual/tutorial/aggregation-with-user-preference-data/）

db.user.aggregate( { $match: { "_id": ObjectId("52426846311fc102641ef332") } } , 
                   { $unwind: "$weightTracker"},  
                   {$group: { _id: null, maxDate: {$max: "$weightTracker" } } }  
                 )