Question

我有这样的文件：

{
    "_id" : ObjectId("53340d07d6429d27e1284c77"),
    "worktypes" : [ 
        {
            "name" : "Pompas",
            "works" : [ 
                {
                    "name" : "work 1",
                    "code" : "0001"
                }
            ]
        },
        {
            "name" : "Pompas "",
            "works" : [ 
                {
                    "name" : "work 2",
                    "code" : "0002"
                }
            ]
        }
    ]
}

我只查询了本文档的工作类型之一的作品，这是查询：

db.categories.find({$and: [
        { "_id": ObjectId('53340d07d6429d27e1284c77')},
        {"worktypes.name": "Pompas"}
    ]},{"worktypes.works.$":1})

但我得到了

    {
    "_id" : ObjectId("53340d07d6429d27e1284c77"),
    "worktypes" : [ 
        {
            "name" : "Pompas",
            "works" : [ 
                {
                    "name" : "work 1",
                    "code" : "0001"
                }
            ]
        }
    ]
}

但我只需要：

"works" : [ 
                    {
                        "name" : "work 1",
                        "code" : "0001"
                    }
                ]

我该如何减少这个？

Answer 1

我认为Neil Lunn's answer大部分是正确的，但在我看来，需要进行一些调整才能获得预期效果：

与"worktypes.name"而不是"worktypes.works.name"
在$group阶段，使用$first代替$push来获取第一个元素
添加$project阶段以获取"works"

db.categories.aggregate([
    { "$unwind": "$worktypes" },
    { "$unwind": "$worktypes.works" },
    { "$match": {
        "worktypes.name": "Pompas"
    }},
    { "$group": {
        "_id": "$_id",
        "works": { "$first": "$worktypes.works" }
    }},
    { "$project": {"_id":0, "works":1} }
])

<强>输出：

{
        "result" : [
                {
                        "works" : {
                                "name" : "work 1",
                                "code" : "0001"
                        }
                }
        ],
        "ok" : 1
}

Answer 2

使用数组时需要使用$unwind运算符：

db.catefories.aggregate([

    // Unwind the first array
    { "$unwind": "$worktypes" },

    // Then unwind the embedded array
    { "$unwind": "$worktypes.works" },

    // Match the item you want
    { "$match": {
        "worktypes.works.name": "work 1"
    }},

    // Group to reform the array structure
    { "$group": {
        "_id": "$_id",
        "worktypes": { "$push": "$worktypes" }
    }}
])

要以$group之后使用$unwind作为数组返回。

Answer 3

您可以选择性地从投影中删除字段，如下所示：

db.categories.find({$and: [
    { "_id": ObjectId('53340d07d6429d27e1284c77')},
    {"worktypes.name": "Pompas"}
]},{"worktypes.works.$":1, _id:0 })

这将阻止投影_id字段。

减少mongoDB查询的响应

3 个答案: