Question

我正在使用MongoDB聚合框架尝试从我们的数据集中收集一些总计。

以下是源数据的示例：

{
    "urn": "urn:content:epi:0001",
    "duration": 3450272,
    "profiles": {
         "low": {
             "mp3": {
                 "size": 8623059425,
                 "url": "0001_low.mp3"
             },
             "oga": {
                 "size": 8623059425,
                 "url": "0001_low.oga"
             },
             "m4a": {
                 "size": 8623059425,
                 "url": "0001_low.m4a"
             }
         },
         "medium": {
             "mp3": {
                 "size": 8623059425,
                 "url": "0001_medium.mp3"
             },
             "oga": {
                 "size": 8623059425,
                 "url": "0001_medium.oga"
             },
             "m4a": {
                 "size": 8623059425,
                 "url": "0001_medium.m4a"
             }
         },
         "high": {
             "mp3": {
                 "size": 8623059425,
                 "url": "0001_high.mp3"
             },
             "oga": {
                 "size": 8623059425,
                 "url": "0001_high.oga"
             },
             "m4a": {
                 "size": 8623059425,
                 "url": "0001_high.m4a"
             }
         }
    }
}

我要做的是将每个profile.(low|medium|high).(mp3|oga|m4a)分成单独的文档/项目进行聚合，例如：

{
    "_id": null,
    "files": [
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_low.mp3"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_low.oga"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_low.m4a"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_medium.mp3"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_medium.oga"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_medium.m4a"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_high.mp3"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_high.oga"
        },
        {
            "urn": "urn:content:epi:0001",
            "duration": 3450272,
            "size": 8623059425,
            "url": "0001_high.m4a"
        }
    ]
}

这种事情是否可以使用聚合框架，或者MapReduce可以做什么？

Answer 1

不，聚合框架的$unwind阶段是将单个集合文档转换为多个管道文档的唯一方法，$unwind仅适用于数组，而不适用于像您这样的子文档。

Mongo将子对象聚合和分离成单独的文档

1 个答案: