查询嵌套数组中元素的总数 - 嵌入文档的MongoDB

Question

我在我的收藏文件喜欢：

{ 
    _id: 1, 
    activities: [ 
    { 
     activity_id: 1, 
     travel: [ 
     { 
      point_id: 1, 
      location: [-76.0,19.1] 
     }, 
     { 
      point_id: 2, 
      location: [-77.0,19.3] 
     } 
     ] 
    }, 
    { 
     activity_id: 2, 
     travel: [ 
     { 
      point_id: 3, 
      location: [-99.3,18.2] 
     } 
     ] 
    } 
    ] 
}, 
{ 
    _id: 2, 
    activities: [ 
    { 
     activity_id: 3, 
     travel: [ 
     { 
      point_id: 4, 
      location: [-75.0,11.1] 
     } 
     ] 
    } 
    ] 
} 

我能得到的活动总数，如下：

db.mycollection.aggregate(
    {$unwind: "$activities"}, 
    {$project: {count:{$add:1}}}, 
    {$group: {_id: null, number: {$sum: "$count" }}} 
) 

我得到（3个活动）：

{ "result" : [ { "_id" : null, "number" : 3 } ], "ok" : 1 } 

问题：何我能获得所有旅行中的元素总数吗？

预期的结果：4元件

它们是：

{ 
    point_id: 1, 
    location: [-76.0,19.1] 
}, 
{ 
    point_id: 2, 
    location: [-77.0,19.3] 
}, 
{ 
    point_id: 3, 
    location: [-99.3,18.2] 
}, 
{ 
    point_id: 4, 
    location: [-75.0,11.1] 
} 

Answer 1

可以很容易地通过使用双$unwind

例如变换文件

db.collection.aggregate([ 
    {$unwind: "$activities"}, 
    {$unwind: "$activities.travel"}, 
    {$group:{ 
    _id:null, 
    travel: {$push: { 
     point_id:"$activities.travel.point_id", 
     location:"$activities.travel.location"}} 
    }}, 
    {$project:{_id:0, travel:"$travel"}} 
]) 

这将发出非常接近你想要的输出格式：

{ 
    "travel" : [ 
     { 
      "point_id" : 1.0, 
      "location" : [ 
       -76.0, 
       19.1 
      ] 
     }, 
     { 
      "point_id" : 2.0, 
      "location" : [ 
       -77.0, 
       19.3 
      ] 
     }, 
     { 
      "point_id" : 3.0, 
      "location" : [ 
       -99.3, 
       18.2 
      ] 
     }, 
     { 
      "point_id" : 4.0, 
      "location" : [ 
       -75.0, 
       11.1 
      ] 
     } 
    ] 
} 

更新：

如果你只是想知道在整个集合旅行证件的总数，

试试看：

db.collection.aggregate([ 
    {$unwind: "$activities"}, 
    {$unwind: "$activities.travel"}, 
    {$group: {_id:0, total:{$sum:1}}} 
]) 

它会打印：

{ 
    "_id" : NumberInt(0), 
    "total" : NumberInt(4) 
} 

更新2：

OP希望基于在聚合框架某些属性过滤文件。这里有一个方法可以这样做：

db.collection.aggregate([ 
    {$unwind: "$activities"}, 
    {$match:{"activities.activity_id":1}}, 
    {$unwind: "$activities.travel"}, 
    {$group: {_id:0, total:{$sum:1}}} 
]) 

它将打印（基于样本文档）：

{ "_id" : 0, "total" : 2 }