Question

我搜索了其他帖子，但是看起来不错的东西在这里无法工作。我需要你的建议。

这是我的文档在数据库中的样子，只有一个文档带有一系列标签。

我只需要查询计数器大于0的餐厅类型（这样最终结果将排除计数器0的任何类型）

我的模式

const tagsSchema = mongoose.Schema({
  _id: mongoose.Schema.Types.ObjectId,
  details: {
    restaurantTypeId: mongoose.Schema.Types.ObjectId,
    restaurantTypes: [{
      _id: mongoose.Schema.Types.ObjectId,
      name: String,
      counter: Number,
    }],
    foodTypeId: mongoose.Schema.Types.ObjectId,
    foodTypes: [{
      _id: mongoose.Schema.Types.ObjectId,
      name: String,
      counter: Number,
    }]
  }
});

我尝试过

    tags.find({
    'details.restaurantTypes.counter': {
        $gt: 0
    }
}, (err, data) => {
    if (err) {
        res.send(err);
    }
    res.json(data);
});

我知道了

[
{
    "details": {
        "restaurantTypeId": "5c01fb57497a896d50f49877",
        "restaurantTypes": [
            {
                "_id": "5c01fb57497a896d50f49879",
                "name": "Asian",
                "counter": 1
            },
            {
                "_id": "5c01fb57497a896d50f4987a",
                "name": "Bakery",
                "counter": 0
            },
            {
                "_id": "5c01fb57497a896d50f4987b",
                "name": "Barbecue",
                "counter": 0
            },
            {
                "_id": "5c01fb57497a896d50f4987c",
                "name": "Bars & Pubs",
                "counter": 0
            },
            {
                "_id": "5c01fb57497a896d50f4987d",
                "name": "Bistro",
                "counter": 0
            },

和

    tags.find({
    'details.restaurantTypes.counter': {
        $gte: 1
    }
}, (err, data) => {
    if (err) {
        res.send(err);
    }
    res.json(data);
});

给我相同的结果

Answer 1

您可以使用聚合管道来过滤restaurantTypes

$match-过滤餐厅
$addFields-通过计数器覆盖restaurantTypes和$filter餐馆类型

汇总管道

 db.res.aggregate([
        {$match: {"_id" : ObjectId("5c2187be640edfe094a3b946")}},
        {$addFields:{"restaurantTypes" : {$filter : {input : "$restaurantTypes", as : "t", cond : {$ne : ["$$t.counter",0]}}}}}
    ])

Answer 2

好的，我找到了答案，灵感来自萨拉瓦那。

这是使用汇总和过滤器的答案。

    tags.aggregate([{
        $match: {
            "_id": mongoose.Types.ObjectId(id)
        }
    },
    {
        $project: {
            "details.restaurantTypes": {
                $filter: {
                    input: "$details.restaurantTypes",
                    as: "resType",
                    cond: {
                        $ne: ["$$resType.counter", 0]
                    }
                }
            }
        }
    }
]

这会给我结果

[
{
    "_id": "5c01fb57497a896d50f49876",
    "details": {
        "restaurantTypes": [
            {
                "_id": "5c01fb57497a896d50f49879",
                "name": "Asian",
                "counter": 1
            },
            {
                "_id": "5c01fb57497a896d50f498a6",
                "name": "Thai",
                "counter": 1
            },
            {
                "_id": "5c01fb57497a896d50f498a8",
                "name": "Western",
                "counter": 1
            }
        ]
    }
}

]

MongoDB查询子文档$ gt

2 个答案: