在 mongodb 中聚合嵌套数组答案

【问题标题】：Aggregating nested arrays in mongodb在 mongodb 中聚合嵌套数组
【发布时间】：2021-06-27 11:14:23
【问题描述】：

我知道我可以使用 $in 来匹配数组中的元素，但是如果数组是嵌套的呢？像这样：

{
    "_id": ObjectId("somethingsomething"),
    "supermarkets": [
        {
            "groceries": [
                {
                    "groceryType": "banana",
                    "groceryStockDate": "12345678",
                    "groceryAmount": 12
                },
                {
                    "groceryType": "cabbage",
                    "groceryStockDate": "313512",
                    "groceryAmount": 53
                },
                {
                    "groceryType": "strawberry",
                    "groceryStockDate": "51362",
                    "groceryAmount": 52
                }
            ]
        },
        {
            "groceries": [
                {
                    "groceryType": "banana",
                    "groceryStockDate": "31321",
                    "groceryAmount": 52
                },
                {
                    "groceryType": "banana",
                    "groceryStockDate": "532451",
                    "groceryAmount": 73
                },
                {
                    "groceryType": "cucumber",
                    "groceryStockDate": "123",
                    "groceryAmount": 12
                }
            ]
        }
    ]
}

在这里，我想用groceryType: banana 获取每个对象，所以最终结果应该是这样的

[
    {
        "groceryType": "banana",
        "groceryStockDate": "12345678",
        "groceryAmount": 12
    },
    {
        "groceryType": "banana",
        "groceryStockDate": "31321",
        "groceryAmount": 52
    },
    {
        "groceryType": "banana",
        "groceryStockDate": "532451",
        "groceryAmount": 73
    }
]

我特别想用聚合来做到这一点，因为我需要稍后通过更多阶段。

【问题讨论】：

标签： mongodb mongodb-query aggregation-framework aggregate aggregation

【解决方案1】：

演示 - https://mongoplayground.net/p/ayerR4qQMoI

在supermarkets 和supermarkets.groceries 上使用$unwind 获取个人文档，使用$match 过滤数据，使用$project 获取正确形状。

db.collection.aggregate({
  $unwind: "$supermarkets"
},
{
  $unwind: "$supermarkets.groceries"
},
{
  $match: {
    "supermarkets.groceries.groceryType": "banana"
  }
},
{
  $project: {
    _id: 0,
    groceries: "$supermarkets.groceries"
  }
})

【讨论】：

@ReturnCos 很高兴它有帮助:)