仅选择子文档或数组

Question

{
    "_id":{
        "oid":"4f33bf69873dbc73a7d21dc3"
    },
    "country":"IND",
    "states":[{
            "name":"orissa",
            "direction":"east",
            "population":41947358,
            "districts":[{
                    "name":"puri",
                    "headquarter":"puri",
                    "population":1498604
                },
                {
                    "name":"khordha",
                    "headquarter":"bhubaneswar",
                    "population":1874405
                }
            ]
        },
        {
            "name":"andhra pradesh",
            "direction":"south",
            "population":84665533,
            "districts":[{
                    "name":"rangareddi",
                    "headquarter":"hyderabad",
                    "population":3506670
                },
                {
                    "name":"vishakhapatnam",
                    "headquarter":"vishakhapatnam",
                    "population":3789823
                }
            ]
        }
    ]
}

在上述集合（即国家/地区）中，我只有一个文档，我想获取有关特定州的详细信息（比如说“country.states.name”：“orissa”），但我希望我的结果在这里在而不是整个文档下。Mogo中有没有办法...

     {
    "name": "orissa",
    "direction": "east",
    "population": 41947358,
    "districts": [
        {
            "name": "puri",
            "headquarter": "puri",
            "population": 1498604
        },
        {
            "name": "khordha",
            "headquarter": "bhubaneswar",
            "population": 1874405
        }
    ]
   }

谢谢

Answer 1

如果您不想使用aggregate，您可以在应用层使用下划线（默认包含）非常轻松地做到这一点：

var country = Groops.findOne({"property":value);
var state _.where(country, {"state":statename});

这将为您提供与statename 匹配的整个状态记录。很方便。

Answer 2

db.countries.find({ "states":  { "$elemMatch": { "name": orissa }}},{"country" : 1, "states.$": 1 })

Answer 3

试过这个：

db.countries.aggregate(      
    {
        "$project": {
            "state": "$states",
            "_id": 0
        }
    },
    {
        "$unwind": "$state"
    },
    {
        "$group": {
            "_id": "$state.name",
            "state": {
                "$first": "$state"
            }
        }
    },
    {
        "$match": {
            "_id": "orissa"
        }
    }
);

得到：

{
    "result" : [
            {
                    "_id" : "orissa",
                    "state" : {
                            "name" : "orissa",
                            "direction" : "east",
                            "population" : 41947358,
                            "districts" : [
                                    {
                                            "name" : "puri",
                                            "headquarter" : "puri",
                                            "population" : 1498604
                                    },
                                    {
                                            "name" : "khordha",
                                            "headquarter" : "bhubaneswar",
                                            "population" : 1874405
                                    }
                            ]
                    }
            }
    ],
    "ok" : 1

Answer 4

你现在不能这样做，但你可以在aggregation framework 中使用 $unwind。你现在可以试试 2.1 的实验分支，稳定版会在 2.2 出来，大概几个月后。

Answer 5

mongodb 中的任何查询总是返回根文档。

如果您知道嵌套数组中的状态序数，则只有一种方法可以通过 $slice 加载一个带有父文档的子文档：

// skip ordinalNumberOfState -1, limit 1
db.countries.find({_id: 1}, {states:{$slice: [ordinalNumber -1 , 1]}}) 

$slice 以默认顺序工作（因为文档被插入到嵌套数组中）。 此外，如果您不需要某个国家/地区的字段，则可以在结果中仅包含 _id 和 states：

db.countries.find({_id: 1}, {states:{$slice: [ordinalNumber -1 , 1]}, _id: 1}) 

然后结果文档将如下所示：

{
    "_id":{
        "oid":"4f33bf69873dbc73a7d21dc3"
    },
    "states":[{
            "name":"orissa",
            "direction":"east",
            "population":41947358,
            "districts":[{
                    "name":"puri",
                    "headquarter":"puri",
                    "population":1498604
                },
                {
                    "name":"khordha",
                    "headquarter":"bhubaneswar",
                    "population":1874405
                }
            ]
        }]
}