Mongoose.aggregate（管道）link 使用 $unwind、$lookup、$group 的多个集合
Mongoose.aggregate(pipeline) link multiple collections using $unwind, $lookup, $group
我是 mongodb 和 mongoose 的聚合功能的新手，在通过我的管道传递数据后一直难以获得所需的结果。
下面我使用虚构的示例模型简化了
场景
我有 3 个模型（Ship、Yatch 和 Sailboat）共享接口并从基础 class 扩展而来。第 4 个模型 Captain，它有一个数组 watercraftContexts，其中包含用于引用与每个 Captain.
关联的 watercrafts 类型的对象
示例 Mongo Data/Schema 设置
// Model Name: 'Captain', Collection Name: 'captains'
{
name: 'Jack Sparrow', // Captian name
license: 'SYS-123', // License Number
classes: ['sail', 'yatch', 'ship'], // Array of allowed operational vessel types
watercraftContexts: [
{
_id: ObjectId('xxxxxxxxxxxxxxx'), // Foreign Model ID
type: 'Sailboat', // Model Name
ref: 'sailboats'. // Collection Name
},
{
_id: ObjectId('xxxxxxxxxxxxxxx'), // Foreign Model ID
type: 'Yatch', // Model Name
ref: 'yatches'. // Collection Name
},
{
_id: ObjectId('xxxxxxxxxxxxxxx'), // Foreign Model ID
type: 'Ship', // Model Name
ref: 'ships'. // Collection Name
}
]
}
如您所见，对象数组已设置为使用 mongoose.populate() 方法和 ref 和 _id 字段，并且我实现了 virtual getter watercrafts 使用 populate() 功能进行保湿（代码未发布）。
当使用 mongoose.Model 查询时，一个新字段被创建为 watercrafts，其中包含来自 3 个不同关联集合的所有对象的数组。
问题
我还需要一种方法来 aggregate 针对此数据产生类似的结果，因为模型方法在 aggregate pipline.
中不可用
这是从我的程序化 mongo 聚合生成的查询：
[ { '$match':
{ _id:
{ '$in':
[ ObjectId('5f77bc653887221a703415e1'),
ObjectId('5f77bc653887221a703415df'),
ObjectId('5f77bc653887221a703415e0'),
ObjectId('5f77bc653887221a703415e5') ] } } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'ships',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.ships' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'yatches',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.yatches' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'sailboats',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.sailboats' } },
{ '$group':
{ _id: '$_id',
watercrafts:
{ '$addToSet':
{ '$concatArrays':
[ '$watercrafts.ships',
'$watercrafts.yatches',
'$watercrafts.sailboats' ] } }
我正在构建一个 mongoose 聚合，如下所示：
const Captain = mongoose.model('Captain')
const aggregate = Captain.aggregrate()
// Dynamically create Aggregate Pipeline in another function
const captains = await Captain.find({})
const captainIds = captains.map(capt => capt._id)
// Match sub-set of documents (in actual project)
aggregate.match({ _id: { $in: captainIds } })
// Collection names to apply $lookup aggregate
const collectionNames = ['sailboats', 'yatches', 'ships']
// Unwind and Lookup for each polymorphic child class's collection
collectionNames.forEach(collection => {
// Separate watercraftContexts into individual records for lookup
aggregate.unwind('watercraftContexts')
// Inner Join collection data on record
aggregate.lookup({
from: collection,
localField: '$watercrafContexts._id',
foreignField: '_id',
// Object keyed by collection name with array of collection records
// to avoid overwrite of previous collection aggregate lookup
as: `watercrafts.${collection}`
})
})
// Re-group the records by Captain Object Id
const aggregateAssociationPaths = collectionNames.map(collection =>
// Mongo Path to each collection $lookup
`$watercrafts.${collection}`
)
// Re-assemble $unwind and $group by Captain's ObjectId
aggregate.group({
_id: '$_id',
$addToSet: {
//
$concatArrays: aggregateAssociationPaths
}
})
/*** !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! ***
* *
* WHAT DO I DO NEXT TO GET ALL THE CAPTAIN DATA WITH THE AGGREGATED `watercrafts`
* *
*** !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! ***/
// Execute Aggregation
const captiansWithWatercraftsAssociations = await aggregate
到目前为止，我的数据看起来像这样，并且该小组没有使用 mongoose:
[ { _id: 5f77bc653887221a703415df,
watercrafts:
[ { _id: 5f77bc653887221a703415d3,
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e0,
watercrafts:
[ { _id: 5f77bc653887221a703415d4,
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e1,
watercrafts:
[ { _id: 5f77bc653887221a703415d5,
class: 'ship',
name: 'Jenny',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e5,
watercrafts:
[ { _id: 5f77bc653887221a703415dd,
class: 'yatch',
name: 'Audrey',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e5,
watercrafts:
[ { _id: 5f77bc653887221a703415dc,
class: 'sail',
name: 'Swell Shredder',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e5,
watercrafts:
[ { _id: 5f77bc653887221a703415de,
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] } ]
感谢支持
对于 MongoDb 的 aggregate 新手来说，这是一个棘手的问题。我会将我的答案分解为多个步骤，以向其他尝试通过引用多个集合来聚合数组的人进行演示。
第 1 步 - $匹配以过滤集合
$match 接受与 db.collection.find({}) 相同的查询，并且 return 是匹配结果数组，在下面的例子中，我 select 4 条特定的记录在这里
{ '$match':
{ _id:
{ '$in':
[
ObjectId('5f7bdb3eea134b5a5c976285'),
ObjectId('5f7bdb3eea134b5a5c976283'),
ObjectId('5f7bdb3eea134b5a5c976284'),
ObjectId('5f7bdb3eea134b5a5c976289')
]
}
}
}
$匹配结果
[
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724d'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e47241'),
watercraftType: 'Sailboat',
ref: 'sailboats' } ],
__v: 0 },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724e'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e47242'),
watercraftType: 'Yatch',
ref: 'yatches' } ],
__v: 0 },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724f'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e47243'),
watercraftType: 'Ship',
ref: 'ships' } ],
__v: 0 },
{ _id: ObjectId('5f7be0b37e2bdf5b19e47253'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e4724a'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724b'),
watercraftType: 'Yatch',
ref: 'yatches' },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724c'),
watercraftType: 'Ship',
ref: 'ships' } ],
__v: 0 }
]
第 2 步 - $unwind，因此我们可以使用 $loopup
进行迭代
在这个结果集中有一个对象数组 { _id: <ObjectId>, watercraftType: <ModelName> } 遍历数组并将这些对象中的每一个与各自的集合记录连接起来，我们必须将数组分解成单独的独立记录。 $unwind 功能将为下一个聚合阶段创建一个新数据集
{ '$unwind': '$watercraftContexts' },
$展开结果
如您所见，$unwind 现在使用单个 watercraftContext 创建一条记录，我们现在设置为使用 $lookup
[ { _id: ObjectId('5f7be2231da37c5b5915bf9b'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf8f'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bf9c'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf90'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bf9d'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf91'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bfa1'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf98'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bfa1'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf99'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bfa1'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf9a'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 } ]
第 4 步 $lookup - 加入外部集合中的每条记录
重要的是要注意，我们必须 $unwind 在为我们需要加入的每个不同集合调用 $lookup 之前。由于我们想要连接多个集合，因此我们需要将结果存储在一个由集合作为键的对象中以供以后聚合。
// Only performs $lookup on 'ships' collection
{ '$lookup':
{ from: 'ships', // Collection Name - Note: repeat for each collection
localField: 'watercraftContexts._id', // The field with id to link
foreignField: '_id', // The field on the foreign collection to match
as: 'watercrafts.ships' // The path where to store the lookup result
}
}
第 5 步 - 对其他联接重复 $unwind 和 $lookup
对其他连接重复上述步骤，并按集合名称键入。我结合了聚合阶段来演示重复。
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'yatches',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.yatches' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'sailboats',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.sailboats' } }
第 4 步和第 5 步的结果
如果仔细观察，您会注意到其中一条 Captain 记录存在 3 次，但另一条 watercraftType 存在。 $lookup 只会 return 记录匹配特定的集合名称。这就是为什么将它们存储在由 collectionName
键入的 Object 中的原因
[
{ _id: ObjectId('5f7be7145320a65b942bb450'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb444'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
{ ships: [],
yatches: [],
sailboats:
[ { _id: ObjectId('5f7be7145320a65b942bb444'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] } },
{ _id: ObjectId('5f7be7145320a65b942bb451'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb445'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
{ ships: [],
yatches:
[ { _id: ObjectId('5f7be7145320a65b942bb445'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ],
sailboats: [] } },
{ _id: ObjectId('5f7be7145320a65b942bb452'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb446'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
{ ships:
[ { _id: ObjectId('5f7be7145320a65b942bb446'),
class: 'ship',
name: 'Jenny',
__v: 0 } ],
yatches: [],
sailboats: [] } },
{ _id: ObjectId('5f7be7145320a65b942bb456'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb44d'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
{ ships: [],
yatches: [],
sailboats:
[ { _id: ObjectId('5f7be7145320a65b942bb44d'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 } ] } },
{ _id: ObjectId('5f7be7145320a65b942bb456'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb44e'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
{ ships: [],
yatches:
[ { _id: ObjectId('5f7be7145320a65b942bb44e'),
class: 'yatch',
name: 'Audrey',
__v: 0 } ],
sailboats: [] } },
{ _id: ObjectId('5f7be7145320a65b942bb456'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb44f'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
{ ships:
[ { _id: ObjectId('5f7be7145320a65b942bb44f'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ],
yatches: [],
sailboats: [] } } ]
第 6 步 $project - 使用 project 来展平连接的对象图
我们可以使用 project select 所有现有数据并将连接结果的 Object Map 扁平化为单个数组。
{ '$project':
// keys with the value 'true' will be included
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts: // Re-assigns value of watercrafts
{ '$setUnion': // Accepts an array of arrays to flatten
[
'$watercrafts.ships',
'$watercrafts.yatches',
'$watercrafts.sailboats'
]
}
}
}
$项目结果
上面$project的结果会将watercrafts对象替换为watercrafts的扁平化数组，但需要注意的是[=]仍然存在重复记录34=] 匹配许多不同的查找。我们将在下一步中 re-piece 他们一起。
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9695'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9689'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9689'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9696'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb968a'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb968a'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9697'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb968b'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb968b'),
class: 'ship',
name: 'Jenny',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb969b'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9692'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9692'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb969b'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9693'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9693'),
class: 'yatch',
name: 'Audrey',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb969b'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9694'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9694'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] } ]
第 7 步 $unwind 和 $group
我们 $unwind 这样我们现在可以将属于同一个 Captain 的所有 watercrafts 分组。我们还必须使用 $mergeObjects 将 Captain 集合中的附加数据临时存储在一个新的临时变量下，为最后阶段做准备。
{ '$unwind': '$watercrafts' },
{ '$group':
{ _id: '$_id',
data:
{ '$mergeObjects':
{ name: '$name',
license: '$license',
classes: '$classes',
watercraftContexts: '$watercraftContexts',
__v: '$__v' } },
watercrafts: { '$push': '$watercrafts' } } }
$unwind 和 $group 结果
现在我们真的有所进展。我们已将转换减少到最初的 4 Captain 并将我们的连接展平为单个数组。
[ { _id: ObjectId('5f7bed5e271dd95c306c25a4'),
data:
{ name: 'CAPTAIN_SHIP',
license: 'WC-3',
classes: [ 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bed5e271dd95c306c2598'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c2598'),
class: 'ship',
name: 'Jenny',
__v: 0 } ] },
{ _id: ObjectId('5f7bed5e271dd95c306c25a8'),
data:
{ name: 'CAPTAIN_SAIL_YATCH_SHIP',
license: 'WC-7',
classes: [ 'sail', 'yatch', 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bed5e271dd95c306c25a1'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c259f'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 },
{ _id: ObjectId('5f7bed5e271dd95c306c25a0'),
class: 'yatch',
name: 'Audrey',
__v: 0 },
{ _id: ObjectId('5f7bed5e271dd95c306c25a1'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] },
{ _id: ObjectId('5f7bed5e271dd95c306c25a2'),
data:
{ name: 'CAPTAIN_SAIL',
license: 'WC-1',
classes: [ 'sail' ],
watercraftContexts:
{ _id: Object('5f7bed5e271dd95c306c2596'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c2596'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ _id: ObjectId('5f7bed5e271dd95c306c25a3'),
data:
{ name: 'CAPTAIN_YATCH',
license: 'WC-2',
classes: [ 'yatch' ],
watercraftContexts:
{ _id: ObjectId('5f7bed5e271dd95c306c2597'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c2597'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] } ]
步骤 8 $replaceRoot 和 $project
我们剩下的就是将我们的 data 合并到每条记录的根目录中，并删除临时变量 data
// Merges 'data' into the root of each record
{ '$replaceRoot': { newRoot: { '$mergeObjects': [ '$data', '$$ROOT' ] } } },
// Use $project to remove data (include only the fields we want)
{ '$project':
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts: true }
}
$replaceRoot & $project 结果
现在我们得到了我们为...A Captain 和混合关联类型数组 watercrafts
设定的结果
[
{ name: 'CAPTAIN_SAIL_YATCH_SHIP',
license: 'WC-7',
classes: [ 'sail', 'yatch', 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755ea6'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ead'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755ea4'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 },
{ _id: ObjectId('5f7bf3b3680b375ca1755ea5'),
class: 'yatch',
name: 'Audrey',
__v: 0 },
{ _id: ObjectId('5f7bf3b3680b375ca1755ea6'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] },
{ name: 'CAPTAIN_SAIL',
license: 'WC-1',
classes: [ 'sail' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755e9b'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ea7'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755e9b'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ name: 'CAPTAIN_YATCH',
license: 'WC-2',
classes: [ 'yatch' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755e9c'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ea8'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755e9c'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] },
{ name: 'CAPTAIN_SHIP',
license: 'WC-3',
classes: [ 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755e9d'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ea9'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755e9d'),
class: 'ship',
name: 'Jenny',
__v: 0 } ] } ]
你已经知道了...只花了 2 天时间就弄明白了。如果您正在尝试类似的聚合关联，我希望它能为您节省一些时间。编码愉快！
最终管道
[
{ '$match':
{ _id:
{ '$in':
[ ObjectId('5f7bf3b3680b375ca1755ea9'),
ObjectId('5f7bf3b3680b375ca1755ea7'),
ObjectId('5f7bf3b3680b375ca1755ea8'),
ObjectId('5f7bf3b3680b375ca1755ead')
]
}
}
},
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'ships',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.ships' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'yatches',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.yatches' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'sailboats',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.sailboats' } },
{ '$project':
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts:
{ '$setUnion':
[ '$watercrafts.ships',
'$watercrafts.yatches',
'$watercrafts.sailboats' ] } } },
{ '$unwind': '$watercrafts' },
{ '$group':
{ _id: '$_id',
data:
{ '$mergeObjects':
{ name: '$name',
license: '$license',
classes: '$classes',
watercraftContexts: '$watercraftContexts',
__v: '$__v' } },
watercrafts: { '$push': '$watercrafts' } } },
{ '$replaceRoot': { newRoot: { '$mergeObjects': [ '$data', '$$ROOT' ] } } },
{ '$project':
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts: true } }
]
我是 mongodb 和 mongoose 的聚合功能的新手，在通过我的管道传递数据后一直难以获得所需的结果。
下面我使用虚构的示例模型简化了
场景
我有 3 个模型（Ship、Yatch 和 Sailboat）共享接口并从基础 class 扩展而来。第 4 个模型 Captain，它有一个数组 watercraftContexts，其中包含用于引用与每个 Captain.
watercrafts 类型的对象
示例 Mongo Data/Schema 设置
// Model Name: 'Captain', Collection Name: 'captains'
{
name: 'Jack Sparrow', // Captian name
license: 'SYS-123', // License Number
classes: ['sail', 'yatch', 'ship'], // Array of allowed operational vessel types
watercraftContexts: [
{
_id: ObjectId('xxxxxxxxxxxxxxx'), // Foreign Model ID
type: 'Sailboat', // Model Name
ref: 'sailboats'. // Collection Name
},
{
_id: ObjectId('xxxxxxxxxxxxxxx'), // Foreign Model ID
type: 'Yatch', // Model Name
ref: 'yatches'. // Collection Name
},
{
_id: ObjectId('xxxxxxxxxxxxxxx'), // Foreign Model ID
type: 'Ship', // Model Name
ref: 'ships'. // Collection Name
}
]
}
如您所见，对象数组已设置为使用 mongoose.populate() 方法和 ref 和 _id 字段，并且我实现了 virtual getter watercrafts 使用 populate() 功能进行保湿（代码未发布）。
当使用 mongoose.Model 查询时，一个新字段被创建为 watercrafts，其中包含来自 3 个不同关联集合的所有对象的数组。
问题
我还需要一种方法来 aggregate 针对此数据产生类似的结果，因为模型方法在 aggregate pipline.
这是从我的程序化 mongo 聚合生成的查询：
[ { '$match':
{ _id:
{ '$in':
[ ObjectId('5f77bc653887221a703415e1'),
ObjectId('5f77bc653887221a703415df'),
ObjectId('5f77bc653887221a703415e0'),
ObjectId('5f77bc653887221a703415e5') ] } } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'ships',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.ships' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'yatches',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.yatches' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'sailboats',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.sailboats' } },
{ '$group':
{ _id: '$_id',
watercrafts:
{ '$addToSet':
{ '$concatArrays':
[ '$watercrafts.ships',
'$watercrafts.yatches',
'$watercrafts.sailboats' ] } }
我正在构建一个 mongoose 聚合，如下所示：
const Captain = mongoose.model('Captain')
const aggregate = Captain.aggregrate()
// Dynamically create Aggregate Pipeline in another function
const captains = await Captain.find({})
const captainIds = captains.map(capt => capt._id)
// Match sub-set of documents (in actual project)
aggregate.match({ _id: { $in: captainIds } })
// Collection names to apply $lookup aggregate
const collectionNames = ['sailboats', 'yatches', 'ships']
// Unwind and Lookup for each polymorphic child class's collection
collectionNames.forEach(collection => {
// Separate watercraftContexts into individual records for lookup
aggregate.unwind('watercraftContexts')
// Inner Join collection data on record
aggregate.lookup({
from: collection,
localField: '$watercrafContexts._id',
foreignField: '_id',
// Object keyed by collection name with array of collection records
// to avoid overwrite of previous collection aggregate lookup
as: `watercrafts.${collection}`
})
})
// Re-group the records by Captain Object Id
const aggregateAssociationPaths = collectionNames.map(collection =>
// Mongo Path to each collection $lookup
`$watercrafts.${collection}`
)
// Re-assemble $unwind and $group by Captain's ObjectId
aggregate.group({
_id: '$_id',
$addToSet: {
//
$concatArrays: aggregateAssociationPaths
}
})
/*** !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! ***
* *
* WHAT DO I DO NEXT TO GET ALL THE CAPTAIN DATA WITH THE AGGREGATED `watercrafts`
* *
*** !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! !!! ***/
// Execute Aggregation
const captiansWithWatercraftsAssociations = await aggregate
到目前为止，我的数据看起来像这样，并且该小组没有使用 mongoose:
[ { _id: 5f77bc653887221a703415df,
watercrafts:
[ { _id: 5f77bc653887221a703415d3,
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e0,
watercrafts:
[ { _id: 5f77bc653887221a703415d4,
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e1,
watercrafts:
[ { _id: 5f77bc653887221a703415d5,
class: 'ship',
name: 'Jenny',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e5,
watercrafts:
[ { _id: 5f77bc653887221a703415dd,
class: 'yatch',
name: 'Audrey',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e5,
watercrafts:
[ { _id: 5f77bc653887221a703415dc,
class: 'sail',
name: 'Swell Shredder',
__v: 0 } ] },
{ _id: 5f77bc653887221a703415e5,
watercrafts:
[ { _id: 5f77bc653887221a703415de,
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] } ]
感谢支持
对于 MongoDb 的 aggregate 新手来说，这是一个棘手的问题。我会将我的答案分解为多个步骤，以向其他尝试通过引用多个集合来聚合数组的人进行演示。
第 1 步 - $匹配以过滤集合
$match 接受与 db.collection.find({}) 相同的查询，并且 return 是匹配结果数组，在下面的例子中，我 select 4 条特定的记录在这里
{ '$match':
{ _id:
{ '$in':
[
ObjectId('5f7bdb3eea134b5a5c976285'),
ObjectId('5f7bdb3eea134b5a5c976283'),
ObjectId('5f7bdb3eea134b5a5c976284'),
ObjectId('5f7bdb3eea134b5a5c976289')
]
}
}
}
[
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724d'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e47241'),
watercraftType: 'Sailboat',
ref: 'sailboats' } ],
__v: 0 },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724e'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e47242'),
watercraftType: 'Yatch',
ref: 'yatches' } ],
__v: 0 },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724f'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e47243'),
watercraftType: 'Ship',
ref: 'ships' } ],
__v: 0 },
{ _id: ObjectId('5f7be0b37e2bdf5b19e47253'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
[ { _id: ObjectId('5f7be0b37e2bdf5b19e4724a'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724b'),
watercraftType: 'Yatch',
ref: 'yatches' },
{ _id: ObjectId('5f7be0b37e2bdf5b19e4724c'),
watercraftType: 'Ship',
ref: 'ships' } ],
__v: 0 }
]
第 2 步 - $unwind，因此我们可以使用 $loopup
进行迭代在这个结果集中有一个对象数组 { _id: <ObjectId>, watercraftType: <ModelName> } 遍历数组并将这些对象中的每一个与各自的集合记录连接起来，我们必须将数组分解成单独的独立记录。 $unwind 功能将为下一个聚合阶段创建一个新数据集
{ '$unwind': '$watercraftContexts' },
如您所见，$unwind 现在使用单个 watercraftContext 创建一条记录，我们现在设置为使用 $lookup
[ { _id: ObjectId('5f7be2231da37c5b5915bf9b'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf8f'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bf9c'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf90'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bf9d'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf91'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bfa1'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf98'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bfa1'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf99'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0 },
{ _id: ObjectId('5f7be2231da37c5b5915bfa1'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be2231da37c5b5915bf9a'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 } ]
重要的是要注意，我们必须 $unwind 在为我们需要加入的每个不同集合调用 $lookup 之前。由于我们想要连接多个集合，因此我们需要将结果存储在一个由集合作为键的对象中以供以后聚合。
// Only performs $lookup on 'ships' collection
{ '$lookup':
{ from: 'ships', // Collection Name - Note: repeat for each collection
localField: 'watercraftContexts._id', // The field with id to link
foreignField: '_id', // The field on the foreign collection to match
as: 'watercrafts.ships' // The path where to store the lookup result
}
}
第 5 步 - 对其他联接重复 $unwind 和 $lookup
对其他连接重复上述步骤，并按集合名称键入。我结合了聚合阶段来演示重复。
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'yatches',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.yatches' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'sailboats',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.sailboats' } }
如果仔细观察，您会注意到其中一条 Captain 记录存在 3 次，但另一条 watercraftType 存在。 $lookup 只会 return 记录匹配特定的集合名称。这就是为什么将它们存储在由 collectionName
Object 中的原因
[
{ _id: ObjectId('5f7be7145320a65b942bb450'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb444'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
{ ships: [],
yatches: [],
sailboats:
[ { _id: ObjectId('5f7be7145320a65b942bb444'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] } },
{ _id: ObjectId('5f7be7145320a65b942bb451'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb445'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
{ ships: [],
yatches:
[ { _id: ObjectId('5f7be7145320a65b942bb445'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ],
sailboats: [] } },
{ _id: ObjectId('5f7be7145320a65b942bb452'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb446'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
{ ships:
[ { _id: ObjectId('5f7be7145320a65b942bb446'),
class: 'ship',
name: 'Jenny',
__v: 0 } ],
yatches: [],
sailboats: [] } },
{ _id: ObjectId('5f7be7145320a65b942bb456'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb44d'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
{ ships: [],
yatches: [],
sailboats:
[ { _id: ObjectId('5f7be7145320a65b942bb44d'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 } ] } },
{ _id: ObjectId('5f7be7145320a65b942bb456'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb44e'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
{ ships: [],
yatches:
[ { _id: ObjectId('5f7be7145320a65b942bb44e'),
class: 'yatch',
name: 'Audrey',
__v: 0 } ],
sailboats: [] } },
{ _id: ObjectId('5f7be7145320a65b942bb456'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7be7145320a65b942bb44f'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
{ ships:
[ { _id: ObjectId('5f7be7145320a65b942bb44f'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ],
yatches: [],
sailboats: [] } } ]
第 6 步 $project - 使用 project 来展平连接的对象图
我们可以使用 project select 所有现有数据并将连接结果的 Object Map 扁平化为单个数组。
{ '$project':
// keys with the value 'true' will be included
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts: // Re-assigns value of watercrafts
{ '$setUnion': // Accepts an array of arrays to flatten
[
'$watercrafts.ships',
'$watercrafts.yatches',
'$watercrafts.sailboats'
]
}
}
}
上面$project的结果会将watercrafts对象替换为watercrafts的扁平化数组，但需要注意的是[=]仍然存在重复记录34=] 匹配许多不同的查找。我们将在下一步中 re-piece 他们一起。
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9695'),
name: 'CAPTAIN_SAIL',
classes: [ 'sail' ],
license: 'WC-1',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9689'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9689'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9696'),
name: 'CAPTAIN_YATCH',
classes: [ 'yatch' ],
license: 'WC-2',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb968a'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb968a'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9697'),
name: 'CAPTAIN_SHIP',
classes: [ 'ship' ],
license: 'WC-3',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb968b'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb968b'),
class: 'ship',
name: 'Jenny',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb969b'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9692'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9692'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb969b'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9693'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9693'),
class: 'yatch',
name: 'Audrey',
__v: 0 } ] },
{ _id: ObjectId('5f7bea8d79dfe25bf3cb969b'),
name: 'CAPTAIN_SAIL_YATCH_SHIP',
classes: [ 'sail', 'yatch', 'ship' ],
license: 'WC-7',
watercraftContexts:
{ _id: ObjectId('5f7bea8d79dfe25bf3cb9694'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
watercrafts:
[ { _id: ObjectId('5f7bea8d79dfe25bf3cb9694'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] } ]
第 7 步 $unwind 和 $group
我们 $unwind 这样我们现在可以将属于同一个 Captain 的所有 watercrafts 分组。我们还必须使用 $mergeObjects 将 Captain 集合中的附加数据临时存储在一个新的临时变量下，为最后阶段做准备。
{ '$unwind': '$watercrafts' },
{ '$group':
{ _id: '$_id',
data:
{ '$mergeObjects':
{ name: '$name',
license: '$license',
classes: '$classes',
watercraftContexts: '$watercraftContexts',
__v: '$__v' } },
watercrafts: { '$push': '$watercrafts' } } }
$unwind 和 $group 结果
现在我们真的有所进展。我们已将转换减少到最初的 4 Captain 并将我们的连接展平为单个数组。
[ { _id: ObjectId('5f7bed5e271dd95c306c25a4'),
data:
{ name: 'CAPTAIN_SHIP',
license: 'WC-3',
classes: [ 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bed5e271dd95c306c2598'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c2598'),
class: 'ship',
name: 'Jenny',
__v: 0 } ] },
{ _id: ObjectId('5f7bed5e271dd95c306c25a8'),
data:
{ name: 'CAPTAIN_SAIL_YATCH_SHIP',
license: 'WC-7',
classes: [ 'sail', 'yatch', 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bed5e271dd95c306c25a1'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c259f'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 },
{ _id: ObjectId('5f7bed5e271dd95c306c25a0'),
class: 'yatch',
name: 'Audrey',
__v: 0 },
{ _id: ObjectId('5f7bed5e271dd95c306c25a1'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] },
{ _id: ObjectId('5f7bed5e271dd95c306c25a2'),
data:
{ name: 'CAPTAIN_SAIL',
license: 'WC-1',
classes: [ 'sail' ],
watercraftContexts:
{ _id: Object('5f7bed5e271dd95c306c2596'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c2596'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ _id: ObjectId('5f7bed5e271dd95c306c25a3'),
data:
{ name: 'CAPTAIN_YATCH',
license: 'WC-2',
classes: [ 'yatch' ],
watercraftContexts:
{ _id: ObjectId('5f7bed5e271dd95c306c2597'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0 },
watercrafts:
[ { _id: ObjectId('5f7bed5e271dd95c306c2597'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] } ]
步骤 8 $replaceRoot 和 $project
我们剩下的就是将我们的 data 合并到每条记录的根目录中，并删除临时变量 data
// Merges 'data' into the root of each record
{ '$replaceRoot': { newRoot: { '$mergeObjects': [ '$data', '$$ROOT' ] } } },
// Use $project to remove data (include only the fields we want)
{ '$project':
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts: true }
}
现在我们得到了我们为...A Captain 和混合关联类型数组 watercrafts
[
{ name: 'CAPTAIN_SAIL_YATCH_SHIP',
license: 'WC-7',
classes: [ 'sail', 'yatch', 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755ea6'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ead'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755ea4'),
class: 'sail',
name: 'Swell Shredder',
__v: 0 },
{ _id: ObjectId('5f7bf3b3680b375ca1755ea5'),
class: 'yatch',
name: 'Audrey',
__v: 0 },
{ _id: ObjectId('5f7bf3b3680b375ca1755ea6'),
class: 'ship',
name: 'Jenny IV',
__v: 0 } ] },
{ name: 'CAPTAIN_SAIL',
license: 'WC-1',
classes: [ 'sail' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755e9b'),
watercraftType: 'Sailboat',
ref: 'sailboats' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ea7'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755e9b'),
class: 'sail',
name: 'Gone with the Wind',
__v: 0 } ] },
{ name: 'CAPTAIN_YATCH',
license: 'WC-2',
classes: [ 'yatch' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755e9c'),
watercraftType: 'Yatch',
ref: 'yatches' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ea8'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755e9c'),
class: 'yatch',
name: 'Liquid Gold',
__v: 0 } ] },
{ name: 'CAPTAIN_SHIP',
license: 'WC-3',
classes: [ 'ship' ],
watercraftContexts:
{ _id: ObjectId('5f7bf3b3680b375ca1755e9d'),
watercraftType: 'Ship',
ref: 'ships' },
__v: 0,
_id: ObjectId('5f7bf3b3680b375ca1755ea9'),
watercrafts:
[ { _id: ObjectId('5f7bf3b3680b375ca1755e9d'),
class: 'ship',
name: 'Jenny',
__v: 0 } ] } ]
你已经知道了...只花了 2 天时间就弄明白了。如果您正在尝试类似的聚合关联，我希望它能为您节省一些时间。编码愉快！
最终管道
[
{ '$match':
{ _id:
{ '$in':
[ ObjectId('5f7bf3b3680b375ca1755ea9'),
ObjectId('5f7bf3b3680b375ca1755ea7'),
ObjectId('5f7bf3b3680b375ca1755ea8'),
ObjectId('5f7bf3b3680b375ca1755ead')
]
}
}
},
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'ships',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.ships' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'yatches',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.yatches' } },
{ '$unwind': '$watercraftContexts' },
{ '$lookup':
{ from: 'sailboats',
localField: 'watercraftContexts._id',
foreignField: '_id',
as: 'watercrafts.sailboats' } },
{ '$project':
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts:
{ '$setUnion':
[ '$watercrafts.ships',
'$watercrafts.yatches',
'$watercrafts.sailboats' ] } } },
{ '$unwind': '$watercrafts' },
{ '$group':
{ _id: '$_id',
data:
{ '$mergeObjects':
{ name: '$name',
license: '$license',
classes: '$classes',
watercraftContexts: '$watercraftContexts',
__v: '$__v' } },
watercrafts: { '$push': '$watercrafts' } } },
{ '$replaceRoot': { newRoot: { '$mergeObjects': [ '$data', '$$ROOT' ] } } },
{ '$project':
{ name: true,
license: true,
classes: true,
_id: true,
watercraftContexts: true,
__v: true,
watercrafts: true } }
]