Return all fields MongoDB Aggregate

Question

I tried searching on here but couldn't really find what I need. I have documents like this:

{
    appletype:Granny,
    color:Green,
    datePicked:2015-01-26,
    dateRipe:2015-01-24,
    numPicked:3
},

{
    appletype:Granny,
    color:Green,
    datePicked:2015-01-01,
    dateRipe:2014-12-28,
    numPicked:6
}

I would like to return only those apples picked latest, will all fields. I want my query to return me the first document only essentially. When I try to do:

db.collection.aggregate([
    { $match : { "appletype" : "Granny" } },
    { $sort : { "datePicked" : 1 } },
    { $group : { "_id" : { "appletype" : "$appletype" },
        "datePicked" : { $max : "$datePicked" } },
])

It does return me all the apples picked latest, however with only appletype:Granny and datePicked:2015-01-26. I need the remaining fields. I tries using $project and adding all the fields, but it didn't get me what I needed. Also, when I added the other fields to the group, since datePicked is unique, it returned both records.

How can I go about returning all fields, for only the latest datePicked?

Thanks!

Answer 1

From your description, it sounds like you want one document for each of the types of apple in your collection and showing the document with the most recent datePicked value.

Here is an aggregate query for that:

db.collection.aggregate([
  { $sort: { "datePicked": -1 },
  { $group: { _id: "$appletype", color: { $first: "$color" }, datePicked: { $first: "$datePicked" }, dateRipe: { $first: "$dateRipe" }, numPicked: { $first: "$numPicked" } } },
  { $project: { _id: 0, color: 1, datePicked: 1, dateRipe: 1, numPicked: 1, appletype: "$_id" } }
])

But then based on the aggregate query you've written, it looks like you're trying to get this:

db.collection.find({appletype: "Granny"}).sort({datePicked: -1}).limit(1);