Question

我在mongoDb集合中有一组BSON文档，我想根据表单级联子文档进行处理。有数千个重复的id仅由时间戳和类型为subsystem的文档 - ＆gt;区分。 component - ＆gt; subcomponent;它们应该基于source相互依附。这是我目前的样本：

[{
    "id": 1001,
    "type": "subsystem",
    "time": 1454767739,
    "payload": {
        "key": "..."
    }
},
{
    "id": 1001,
    "type": "subsystem",
    "time": 1454767539,
    "payload": {
        "key": "..."
    }
},
{
    "id": 2001,
    "type": "component",
    "time": 1454767778,
    "source": 1001,
    "payload": {
        "key": "..."
    }
},
{
    "id": 5002,
    "type": "subcomponent",
    "time": 1454767779,
    "source": 2001,
    "payload": {
        "key": "..."
    }
},
{
    "id": 5003,
    "type": "subcomponent",
    "time": 1454767798,
    "source": 2001,
    "payload": {
        "key": "..."
    }
}]

我想做的是：

匹配时间戳低于maxTime值的所有元素
按（唯一）ID对所有元素进行分组，但仅使用$max: $time的最新值（$$ROOT）（现在每id都是唯一的）
按属性source将所有内容分组为具有该id（或_id）

我已经实现了1.）和2.）已经有了这个小组：

[{
    "_id": 1001,
    "reftime": 1454767739,
    "element": [
        {
            "id": 1001,
            "type": "subsystem",
            "time": 1454767739,
            "payload": {
                "key": "..."
            }
        }
    ]

},
{
    "_id": 2001,
    "reftime": 1454767778,
    "element": [
        {
            "id": 2001,
            "type": "component",
            "time": 1454767778,
            "source": 1001,
            "payload": {
                "key": "..."
            }
        }

    ]
},
{
    "_id": 5002,
    "reftime": 1454767779,
    "element": [
        {
            "id": 5002,
            "type": "subcomponent",
            "time": 1454767779,
            "source": 2001,
            "payload": {
                "key": "..."
            }
        }
    ]
},
{
    "_id": 5003,
    "reftime": 1454767798,
    "element": [
        {
            "id": 5003,
            "type": "subcomponent",
            "time": 1454767798,
            "source": 2001,
            "payload": {
                "key": "..."
            }
        }
    ]
}]

但我想要的是根据source值中的链接构建的元素。所以它看起来像这样：

{
    "_id": 1001,
    "reftime": 1454767739,
    "element": [
        {
            "id": 1001,
            "type": "subsystem",
            "time": 1454767739,
            "payload": {
                "key": "..."
            }
        }
    ],
    "attached": [
        {
            "_id": 2001,
            "reftime": 1454767778,
            "element": [
                {
                    "id": 2001,
                    "type": "component",
                    "time": 1454767778,
                    "source": 1001,
                    "payload": {
                        "key": "..."
                    }
                }
            ],
            "attached": [
                {
                    "_id": 5002,
                    "reftime": 1454767779,
                    "element": [
                        {
                            "id": 5002,
                            "type": "subcomponent",
                            "time": 1454767779,
                            "source": 2001,
                            "payload": {
                                "key": "..."
                            }
                        }
                    ]
                },
                {
                    "_id": 5003,
                    "reftime": 1454767798,
                    "element": [
                        {
                            "id": 5003,
                            "type": "subcomponent",
                            "time": 1454767798,
                            "source": 2001,
                            "payload": {
                                "key": "..."
                            }
                        }
                    ]
                }                

            ]
        }        
    ]

}

这可能吗？我正在努力组建一个包含source的所有内容的组，并将其与样本2中的组合在一起。）。或者，如果这会有所帮助，我可以介绍source: 0？对于每个subsystem和一个根元素下的组？

任何想法或见解？我现在卡在这一点上。谢谢！

mongoDb：聚合文档对子组的引用

0 个答案: