aggregation framework—在mongoDB中执行union

Question

我想知道如何在MongoDB中执行一种聚合。让我们把下面的文档想象成一个集合(为了示例的方便，使用这个结构):

{
  linkedIn: {
    people : [
    {
      name : 'Fred'
     },
     {
       name : 'Matilda'
     }
   ]
  },
  twitter: {
    people : [
    {
       name : 'Hanna'
    },
    {
       name : 'Walter'
    }
   ]
  }
 }

如何使聚合返回twitter和linkedIn中人员的联合?

{
 { name :'Fred', source : 'LinkedIn'},
 { name :'Matilda', source : 'LinkedIn'},
 { name :'Hanna', source : 'Twitter'},
 { name :'Walter', source : 'Twitter'},
}

Answer 1

有几种方法可以使用

的聚合方法

db.collection.aggregate([
    // Assign an array of constants to each document
    { "$project": {
        "linkedIn": 1,
        "twitter": 1,
        "source": { "$cond": [1, ["linkedIn", "twitter"],0 ] }
    }},
    // Unwind the array
    { "$unwind": "$source" },
    // Conditionally push the fields based on the matching constant
    { "$group": { 
        "_id": "$_id",
        "data": { "$push": {
            "$cond": [
                { "$eq": [ "$source", "linkedIn" ] },
                { "source": "$source", "people": "$linkedIn.people" },
                { "source": "$source", "people": "$twitter.people" }
            ]
        }}
    }},
    // Unwind that array
    { "$unwind": "$data" },
    // Unwind the underlying people array
    { "$unwind": "$data.people" },
    // Project the required fields
    { "$project": {
        "_id": 0,
        "name": "$data.people.name",
        "source": "$data.source"
    }}
])

或者用不同的方法使用MongoDB 2.6中的一些操作符:

db.people.aggregate([
    // Unwind the "linkedIn" people
    { "$unwind": "$linkedIn.people" },
    // Tag their source and re-group the array
    { "$group": {
        "_id": "$_id",
        "linkedIn": { "$push": {
            "name": "$linkedIn.people.name",
            "source": { "$literal": "linkedIn" }
        }},
        "twitter": { "$first": "$twitter" }
    }},
    // Unwind the "twitter" people
    { "$unwind": "$twitter.people" },
    // Tag their source and re-group the array
    { "$group": {
        "_id": "$_id",
        "linkedIn": { "$first": "$linkedIn" },
        "twitter": { "$push": {
            "name":  "$twitter.people.name",
            "source": { "$literal": "twitter" }
        }}
    }},
    // Merge the sets with "$setUnion"
    { "$project": {
        "data": { "$setUnion": [ "$twitter", "$linkedIn" ] }
    }},
    // Unwind the union array
    { "$unwind": "$data" },
    // Project the fields
    { "$project": {
        "_id": 0,
        "name": "$data.name",
        "source": "$data.source"
    }}
])

当然，如果你根本不在乎来源是什么:

db.collection.aggregate([
    // Union the two arrays
    { "$project": {
        "data": { "$setUnion": [
            "$linkedIn.people",
            "$twitter.people"
        ]}
    }},
    // Unwind the union array
    { "$unwind": "$data" },
    // Project the fields
    { "$project": {
        "_id": 0,
        "name": "$data.name",
    }}
])

Answer 2

不确定对于这种操作是否建议使用aggregate而不是map-reduce，但以下是您所要求的(不知道是否可以在.aggregate()函数中使用$const而没有任何问题):

aggregate([ 
   { $project: { linkedIn: '$linkedIn', twitter: '$twitter', idx: { $const: [0,1] }}},
   { $unwind: '$idx' },
   { $group: { _id : '$_id', data: { $push: { $cond:[ {$eq:['$idx', 0]}, { source: {$const: 'LinkedIn'}, people: '$linkedIn.people' } , { source: {$const: 'Twitter'}, people: '$twitter.people' } ] }}}},
   { $unwind: '$data'},
   { $unwind: '$data.people'},
   { $project: { _id: 0, name: '$data.people.name', source: '$data.source' }}
])