Javascript 多聚合查询MongoDB

我有以下情况：

• 收集

  部门

  ：

  { "_id" : 0, "name" : "Dep_A_1", "description" : "Department A1" }
  { "_id" : 1, "name" : "Dep_B_1", "description" : "Department B1" }
  { "_id" : 2, "name" : "Dep_C_1", "description" : "Department C1" }
  

• 收藏

  技能

  ：

  { 
      "_id" : 0, 
      "name" : "Creativity", 
      "description" : "description", 
      "type" : "soft", 
      "categories" : [ 0, 2 ] 
  }
  

• 集合

  角色

  ：

  { 
      "_id" : 0,
      "name" : "Software Developer Automation and Control",
      "description" : "Automation and Control expert",
      "departments" : [ 0, 2 ],
      "skills" : [ { "_id" : 18, "weight" : 30 } ]
  }
  

我需要这样的结果：

{
    "_id" : 0, 
    "name" : "Software Developer Automation and Control", 
    "description" : "Automation and Control expert",
    "departments" : [ 
        { 
            "_id" : 0, 
            "name" : "Dep_A_1",
            "description" : "Department A1" 
        }, 
        { 
            "_id" : 2, 
            "name" : "Dep_C_1", 
            "description" : "Department C1" 
        }
    ], 
    "skills" : [ 
        {
            "_id" : 0, 
            "name" : "Creativity", 
            "description" : "description", 
            "type" : "soft", 
            "weight" : 30
        }
    ]
}

我需要将

角色.departments

和

角色.skills

数组替换为

部门

和

角色

集合中的相应对象。有没有办法查询Mongo并得到这样的结果

无论如何，我使用的是Mongo3.6和Pymango。 谢谢。

只需使用两次：

编辑：现在正确保留了

角色.技能.权重

：

db.getCollection('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'}
    }},
    {$lookup: {
        from: 'departments',
        localField: 'departments',
        foreignField: '_id',
        as: 'departments'
    }}
])

---

为了避免成本高昂的

$REWIND

和

$group

阶段，您可以这样做：

db.roles.aggregate([{
    $lookup: {
        from: 'departments',
        localField: 'departments',
        foreignField: '_id',
        as: 'departments'
    }
}, {
    $lookup: {
        from: 'skills',
        let: {
            "skills": "$skills" // keep a reference to the "skills" field of the current document and make it accessible via "$$skills"
        },
        pipeline: [{
            $match: {
                $expr: {
                    $in: [ "$_id", "$$skills._id" ] // this just does the "joining"
                }
            }
        }, {
            $addFields: { // add a new field
                "weight": { // called "weight"
                    $arrayElemAt: [ // which shall be set to the correct "weight"
                        "$$skills.weight", // that can be found in the "weight" field of the "skills" array
                        { $indexOfArray: [ "$$skills._id", "$_id" ] } // at the position that has the matching "_id" value
                    ]
                }
            }
        }],
        as: 'skills'
    }
}])

---

到目前为止你试过什么？你能展示一些东西吗？这对部门来说非常有效，但这样我就失去了角色、技能和权重的价值。有解决方案吗？您好，您的答案会产生以下错误：“$arrayElemAt的第二个参数必须是数值，但是对象”您确定您使用的是最新版本的代码吗？我想在我后来添加的“indexOfArray”前面最初缺少了一个

$

。它肯定能在我的机器上工作——我只是再试一次。

db.roles.aggregate([{
    $lookup: {
        from: 'departments',
        localField: 'departments',
        foreignField: '_id',
        as: 'departments'
    }
}, {
    $lookup: {
        from: 'skills',
        let: {
            "skills": "$skills" // keep a reference to the "skills" field of the current document and make it accessible via "$$skills"
        },
        pipeline: [{
            $match: {
                $expr: {
                    $in: [ "$_id", "$$skills._id" ] // this just does the "joining"
                }
            }
        }, {
            $addFields: { // add a new field
                "weight": { // called "weight"
                    $arrayElemAt: [ // which shall be set to the correct "weight"
                        "$$skills.weight", // that can be found in the "weight" field of the "skills" array
                        { $indexOfArray: [ "$$skills._id", "$_id" ] } // at the position that has the matching "_id" value
                    ]
                }
            }
        }],
        as: 'skills'
    }
}])