具有多个id的MongoDB组

Question

我有一组文件,每个文件中有20个以上的密钥,密钥因文档而异.某些密钥可能不存在于所有文档中.我正在尝试使用聚合框架运行MongoDB组操作.查询看起来像这样 -

db.collection.aggregate([{'$group': {'count': {'$sum': 1}, '_id': {'location': '$location', 'type': '$type', 'language': '$language'}}}])

在理想情况下,它应该返回应该存在3个键的文档,并对它们执行"分组"操作.但结果看起来像这样 -

{
    "result" : [
        {
            "_id" : {
                "location" : "abc",
                "type" : "456"
            },
            "count" : 5
        },
        {
            "_id" : {
                "type" : "123",
                "language" : "english"
            },
            "count" : 1
        },
        {
            "_id" : {
                "location" : "ghi",
                "type" : "9876",
                "language" : "latin"
            },
            "count" : 2
        },
        {
            "_id" : {
                "language" : "hebrew",
                "type" : "9434"
            },
            "count" : 3
        },
        {
            "_id" : {
                "type" : "525",
                "location" : "cari"
            },
            "count" : 1
        },
        {
            "_id" : {
                "language" : "spanish",
                "location" : "dff"
            },
            "count" : 12
        },
        {
            "_id" : {
                "location" : "adpj",
                "type" : "3463",
                            "language": "english"
            },
            "count" : 8
        },
        {
            "_id" : {
                "language" : "french",
                "location" : "nts"
            },
            "count" : 6
        }
    ],
    "ok" : 1
}

问题是,即使MongoDB找不到我在查询中要求的所有3个键并显示部分分组,它也会进行组操作.我只对获得所有键的结果感兴趣.客户端过滤不是一种选择.任何人都可以帮忙吗？

Answer 1

对于MongoDB的$ group-operator,没有值也是值.

如果要排除不存在所有三个键的任何文档,可以向聚合管道添加$ match -step,过滤任何没有所有这些键的文档.

 db.collection.aggregate([
     { $match: { 
         "type" : { "$exists" : true}, 
         "location" : { "$exists" : true}, 
         "language" : { "$exists" : true}
       } 
     },
     { $group: {
         "_id": {
             "location": "$location", 
             "type": "$typ", 
             "language": "$language"
         },
         "count": {$sum: 1}
       }
     }
 ]);

Answer 2

询问：

db.collection.aggregate([{
    $match: {
        type: {
            "$exists": true
        },
        location: {
            "$exists": true
        },
        language: {
            "$exists": true
        }
    }
}])