Mongodb Group 并从文档中获取其他字段

Question

update so Mohammad Faisal has the best solution.However it breaks when a new document is added lol! so i learned a lot from his code and modified it and it Works! =) the code is all the way in the bottom.

But here's what i said.. So i have this document

{"_id":"5ddea2e44eb407059828d740",
"projectname":"wdym",
"username":"easy",
"likes":0,
"link":["ssss"]
}


{"_id":"5ddea2e44eb407059822d740",
"projectname":"thechosenone",
"username":"easy",
"likes":30,
"link":["ssss"]
}

{"_id":"5ddea2e44eb407059828d740",
"projectname":"thanos",
"username":"wiley",
"likes":10,
"link":["ssss"]
}

and basically what i want is the document that contains the highest likes with it's associated project name

For example the output would be

"projectname":"thechosenone",
"username":"easy",
"likes":30
}
,
{
"projectname":"thanos",
"username":"wiley",
"likes":10,
}

the code i have for this is the following

db
    .collection("projects")
    .aggregate([
      {
        $group: {
          _id: { username: "$username" },
          likes: { $max: "$likes" }
        }
      },
      {
        $project:{projectname:1}
      }
    ])

$project gives me a strange output. However, the output was correct without the $project. But i wanted to project the projectname, the user and the highest likes. Thanks for hearing me out :)

heres the solution =)

db
    .collection("projects")
    .aggregate([
      {
        $sort: {
          likes: -1
        }
      },
      {
        $group: {
          _id: {
            username: "$username"
          },
          likes: {
            $max: "$likes"
          },
          projectname: {
            $push: "$projectname"
          },
          link: {
            $push: "$link"
          }
        }
      },
      {
        $project: {
          username: "$_id.username",
          projectname: {
            $arrayElemAt: ["$projectname", 0]
          },
          link: {
            $arrayElemAt: ["$link", 0]
          }
        }
      }
    ])
    .toArray()

Answer 1

If you don't have to use $group this will solve your problem:

db.projects.aggregate([
  {$sort:{likes:-1}},
  {$limit:1} 
]).pretty()

the result would be

{
  "_id" : ObjectId("5ddee7f63cee7cdf247059db"),
  "projectname" : "thechosenone",
  "username" : "easy",
  "likes" : 30,
  "links" : ["ssss"]
}

Answer 2

Try this:-

db.collection("projects").aggregate([
  {
    $group: {
      _id: { username: "$username" },
      likes: { $max: "$likes" },
      projectname: { $push : { $cond: [  { $max: "$likes" }, "$projectname", "" ]}}
    }
  }
  ,
  {
    $project:{
       username:"$_id.username",
       projectname:{"$reduce": {
            "input": "$projectname",
            "initialValue": { "$arrayElemAt": ["$projectname", 0] },
            "in": { "$cond": [{ "$ne": ["$$this", ""] }, "$$this", "$$value"] }
        }},
       likes:1
    }
  }
])