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Question

我有这样的文件：

{
    "_id" : ObjectId("5d302a6da9b1d6ae926c16a3"),
    "location" : "car",
    "playlist" : [ 
        {
            "playlist_number" : "1",
            "number_of_songs" : "100",
            "drive" : [ 
                {
                   "drive_number" : "13",
                    "drive_name" : "Ooms",
                    "songs" : [ 
                        "59 Volvo", 
                        "Yield Not To Temptation - Single Version", 
                        "Johnny Angel", 
                        "The Madison Time", 
                        "Little Brown Jug", 
                        "Only In My Dreams"
                    ]
                }, 
                 {
                   "drive_number" : "Z3",
                    "drive_name" : "codes",
                    "songs" : [ 
                        "59 Volvo", 
                        "Perfect", 
                        "Hands clap", 
                        "I Ran", 
                        "Falling down", 
                        "Sounds of middle earth"
                    ]
                }
            ]
        }
    ]
}

可能还有许多其他location，它们可能包含数量可变的playlist个对象（而对象又可以包含n个数量的playlist_number，您会发现）。

当我按名称搜索歌曲时，例如：59 Volvo，我应该返回location，playlist_number，drive_number和drive_name。我尝试使用嵌套的$elemMatch，但它给出了整个文档。

Answer 1

在运行$unwind之前，您需要$match数据。然后，您可以使用$project重塑数据。另外，您可以添加一个$match作为第一步，以便在进一步处理之前过滤掉不必要的文档：

db.collection.aggregate([
    {
        $match: { "playlist.drive.songs": "59 Volvo" }
    },
    {
        $unwind: "$playlist"
    },
    {
        $unwind: "$playlist.drive"
    },
    {
        $match: { "playlist.drive.songs": "59 Volvo" }
    },
    {
        $project: {
            location: 1,
            drive_number: "$playlist.drive.drive_number",
            drive_name: "$playlist.drive.drive_name",
            playlist_number: "$playlist.playlist_number",
        }
    }
])

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