Mongodb Group 并从文档中获取其他字段

Question

更新所以 Mohammad Faisal 有最好的解决方案。但是，当添加新文档时它会中断，哈哈！所以我从他的代码中学到了很多东西并对其进行了修改，它可以工作！=) 代码一直在底部。但这就是我所说的......所以我有这个文件{"_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"]}基本上我想要的是包含最高赞及其相关项目名称的文档例如输出将是"projectname":"thechosenone","username":"easy","likes":30},{"projectname":"thanos","username":"wiley","likes":10,}我为此拥有的代码如下db    .collection("projects")    .aggregate([      {        $group: {          _id: { username: "$username" },          likes: { $max: "$likes" }        }      },      {        $project:{projectname:1}      }    ])$project 给了我一个奇怪的输出。但是，如果没有 $project，输出是正确的。但我想投影项目名称、用户和最高点赞数。谢谢你听我说:)

Answer 1

如果您不必使用 $group 这将解决您的问题：

db.projects.aggregate([

  {$sort:{likes:-1}},

  {$limit:1} 

]).pretty()

结果是

{

  "_id" : ObjectId("5ddee7f63cee7cdf247059db"),

  "projectname" : "thechosenone",

  "username" : "easy",

  "likes" : 30,

  "links" : ["ssss"]

}

Answer 2

试试这个：-

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

    }

  }

])