Question

请考虑以下学校的收藏学校：

{
    "_id" : 1,
    "name" : "Class 1",
    "students" : [  
        { "rollNo" : 10001, "name" : "Ram", "score" : 65 },
        { "rollNo" : 10002, "name" : "Shyam", "score" : 90 },
        { "rollNo" : 10003, "name" : "Mohan", "score" : 75 }       
        ]
},
{
    "_id" : 2,
    "name" : "Class 2",
    "students" : [  
        { "rollNo" : 20001, "name" : "Krishna", "score" : 88 },
        { "rollNo" : 20002, "name" : "Sohan", "score" : 91 },
        { "rollNo" : 20003, "name" : "Radhika", "score" : 82 },
        { "rollNo" : 20004, "name" : "Komal", "score" : 55 },
        { "rollNo" : 20005, "name" : "Sonam", "score" : 91 }            
        ]
},
{
    "_id" : 3,
    "name" : "Class 3",
    "students" : [  
        { "rollNo" : 30001, "name" : "Monika", "score" : 77 },      
        { "rollNo" : 30002, "name" : "Rahul", "score" : 81 }                
        ]
}

在这里，我的目标是从每个杂项Orderby降序得分中获取前N名学生（考虑按分数排序的前2名学生）。我的预期结果是：

Answer 1

如果您只希望学生记录而不进行分组，则可以简单地$unwind，然后$sort：

db.collection.aggregate([
  { $unwind: "$students" },
  { $sort: { "students.score": -1 } },
  { $limit: 2 }
])

这将留下students对象。为了获得更清晰的输出，您可以将$replaceRoot与$mergeObjects结合使用：

db.collection.aggregate([
  { $unwind: "$students" },
  { $sort: { "students.score": -1 } },
  { $replaceRoot: {
      newRoot: {
        $mergeObjects: [ { _id: "$_id", class: "$name" }, "$students" ]
      }
    }
  },
  { $limit: 2 }
])

查看working here

这将为您提供以下输出：

[
  {
    "_id": 2,
    "class": "Class 2",
    "name": "Sonam",
    "rollNo": 20005,
    "score": 91
  },
  {
    "_id": 2,
    "class": "Class 2",
    "name": "Sohan",
    "rollNo": 20002,
    "score": 91
  }
]

更新：

使用它来获得每个组的前2名：

db.collection.aggregate([
  { $unwind: "$students" },
  { $sort: { "students.score": -1 }
  },
  {
    $group: {
      "_id": "$_id",
      "name": { $first: "$name" },
      "students": { $push: "$students" }
    }
  },
  {
    "$project": {
      "top_two": { "$slice": [ "$students", 2 ] }
    }
  }
])

查看working here

通过降级MongoDB从每个阵列顺序中选择前N个项目

1 个答案: