聚合 MongoDB 后重命名字段答案

【问题标题】：Rename field after aggregate MongoDB聚合 MongoDB 后重命名字段
【发布时间】：2018-08-06 19:29:19
【问题描述】：

在 MongoDB 中执行以下查询

roles.aggregate([
      {'$unwind': '$skills'},
      {'$lookup': {
          'from': 'skills',
          'localField': 'skills._id',
          'foreignField': '_id',
          'as': 'skillsInfo'
      }},
      {'$unwind': '$skillsInfo'},
      {'$addFields': {'skills': {'$mergeObjects': ['$skills', '$skillsInfo']}}},
      {'$project': {'skillsInfo': 0}},
      {'$group': {
          '_id': '$_id',
          'name': {'$first': '$name'},
          'description': {'$first': '$description'},
          'departments': {'$first': '$departments'},
          'skills': {'$push': '$skills'},
          'deep' : {'$first' : '$deep'},
          'range': {'$first' : '$range'},
          'employees': {'$first' : '$employees'}
      }},
      {'$lookup': {
          'from': 'departments',
          'localField': 'departments',
          'foreignField': '_id',
          'as': 'departments'
      }}
  ])

我有这样的结果：

{
  "code": 10, 
  "deep": 1, 
  "departments": [
    {
      "_id": 0, 
      ....
    }
  ], 
  "description": "ER expert", 
  "employees": [
    1, 
    0
  ], 
  "name": "Database Designer", 
  "range": {
    "int": 75, 
    "min": 40
  }, 
  "skills": [
    {
      "_id": 2, 
      ....
    }
  ]
}

有没有办法在“code”中替换子文档键“_id”的名称？例如，我希望“departments._id”为“departments.code”。技能也是一样。无论如何，如果重要的话，我正在使用 MongoDB 3.6 和 PyMongo。谢谢

【问题讨论】：

您使用的是哪个 MongoDB 版本？

标签： mongodb

【解决方案1】：

从 v3.6 开始，$lookup 阶段支持special syntax for "nested queries"。尝试用以下版本替换最后的 $lookup 阶段：

$lookup: {
    from: "departments",
    let: { "departments": "$departments" },
    pipeline: [{
        $match: {
            $expr: {
                $in: [ "$_id",  "$$departments" ] // this just does the "join"
            }
        }
    }, {
        $addFields: {
            "code": "$_id" // create a field named "code" that contains the "_id" field's value
        }
    }, {
        $project: {
            _id: 0 // remove _id field
        }
    }],
    as: "departments"
}

【讨论】：

【解决方案2】：

您也可以使用$map。通过向您已有的结果添加新阶段：

{
    $project: {
        code: 1,
        deep: 1,
        departments: {
            $map: {
                input: "$departments",
                as: 'department',
                in: {
                        "code" : "$$department._id",
                        "department_field_1" : "$$department.department_field_1",
                        "department_field_2" : "$$department.department_field_2",
                        ...
                }
            }
        },
        description: 1,
        employees: 1,
        name: 1,
        range: 1,
        skills: {
            $map: {
                input: "$skills",
                as: 'skill',
                in: {
                        "code" : "$$skill._id",
                        "skill_field_1" : "$$skill.skill_field_1",
                        "skill_field_2" : "$$skill.skill_field_2",
                        ...
                }
            }
        }
    }
}

【讨论】：

【解决方案3】：

这可能有助于from docs

您可以在哪里使用 $replaceRoot 与嵌套在数组中的文档

例如，如果您有这样的文件：

{
      "_id" : 1,
      "grades" : [
         { "test": 1, "grade" : 80, "mean" : 75, "std" : 6 },
         { "test": 2, "grade" : 85, "mean" : 90, "std" : 4 },
         { "test": 3, "grade" : 95, "mean" : 85, "std" : 6 }
      ]
   },
   {
      "_id" : 2,
      "grades" : [
         { "test": 1, "grade" : 90, "mean" : 75, "std" : 6 },
         { "test": 2, "grade" : 87, "mean" : 90, "std" : 3 },
         { "test": 3, "grade" : 91, "mean" : 85, "std" : 4 }
      ]
   }

以下操作将grade字段大于或等于90的嵌入文档提升到顶层：

db.students.aggregate( [
   { $unwind: "$grades" },
   { $match: { "grades.grade" : { $gte: 90 } } },
   { $replaceRoot: { newRoot: "$grades" } }
] )

给出输出

{ "test" : 3, "grade" : 95, "mean" : 85, "std" : 6 }
{ "test" : 1, "grade" : 90, "mean" : 75, "std" : 6 }
{ "test" : 3, "grade" : 91, "mean" : 85, "std" : 4 }

【讨论】：