mongodb - 如何根据 MongoDB 中的列表计数过滤文档？

Question

在 MongoDB 中，我正在寻找至少包含 5 条餐厅评论的文档。我尝试在 $match 上添加过滤器，例如：

{"grades.score: {$gt: 4}}

但是，这将为我提供评论评分至少为 5 或更高的文档，而我想要的是评论计数至少为 5。

db.restaurants.aggregate([
...         {"$match":
...             {"$and":[
...                 {"borough":"Bronx"},
...                 { "cuisine": "American "},
...                 {"address.zipcode":"10467"}]}},
...     {"$group":
...         {"_id":
...             {"name" : "$name",
...              "grades.score" : "$grades.score"}}}])

这就是我得到的：

{ "_id" : { "name" : "Zymi Bar & Grill", "grades.score" : [ 75, 5, 18 ] } }
{ "_id" : { "name" : "Allerton Diner", "grades.score" : [ 20, 9 ] } }
{ "_id" : { "name" : "Gasolina Bar Lounge", "grades.score" : [ 10, 10 ] } }
{ "_id" : { "name" : "Salud Y Estilo De Vida", "grades.score" : [ 0, 7, 7, 6 ] } }
{ "_id" : { "name" : "Coffee Shop", "grades.score" : [ 20, 12, 12, 9, 2 ] } }
{ "_id" : { "name" : "Nicky'S Coffee Shop", "grades.score" : [ 7, 9, 12, 13, 3 ] } }
{ "_id" : { "name" : "John'S Luncheonette", "grades.score" : [ 5, 5, 8, 12, 11 ] } }
{ "_id" : { "name" : "Kennedy'S Chicken & Pizza", "grades.score" : [ 7 ] } }
{ "_id" : { "name" : "V.I.P.'S Cafe", "grades.score" : [ 9, 11, 17, 7, 23, 9 ] } }
{ "_id" : { "name" : "Woodlawn Cafe", "grades.score" : [ 5, 13, 12, 11, 27, 2 ] } }
{ "_id" : { "name" : "Moshulo Golf Course", "grades.score" : [ 11 ] } }
{ "_id" : { "name" : "Kennedy Fried Chicken", "grades.score" : [ 4, 9, 5, 9 ] } }
{ "_id" : { "name" : "Kennedy'S Chicken And Pizza", "grades.score" : [ 9, 19, 13, 12, 26 ] } }
{ "_id" : { "name" : "502 Bar Lounge", "grades.score" : [ 4, 4, 2, 0 ] } }
{ "_id" : { "name" : "Burger Barn Restaurant", "grades.score" : [ 8, 9, 11, 16, 19 ] } }

这就是我需要的：

{ "_id" : { "name" : "Coffee Shop", "grades.score" : [ 20, 12, 12, 9, 2 ] } }
{ "_id" : { "name" : "Nicky'S Coffee Shop", "grades.score" : [ 7, 9, 12, 13, 3 ] } }
{ "_id" : { "name" : "John'S Luncheonette", "grades.score" : [ 5, 5, 8, 12, 11 ] } }
{ "_id" : { "name" : "V.I.P.'S Cafe", "grades.score" : [ 9, 11, 17, 7, 23, 9 ] } }
{ "_id" : { "name" : "Woodlawn Cafe", "grades.score" : [ 5, 13, 12, 11, 27, 2 ] } }
{ "_id" : { "name" : "Kennedy'S Chicken And Pizza", "grades.score" : [ 9, 19, 13, 12, 26 ] } }
{ "_id" : { "name" : "Burger Barn Restaurant", "grades.score" : [ 8, 9, 11, 16, 19 ] } }

标签： mongodbaggregation-framework

Answer 1

您可以根据您的示例尝试此操作：

db.restaurants.aggregate([{
        "$match": {
            "$and": [{
                    "borough": "Bronx"
                },
                {
                    "cuisine": "American "
                },
                {
                    "addresszipcode": "10467"
                }
            ]
        }
    },
    {
        "$group": {
            "_id": {
                "name": "$name",
                "score": "$grades.score",
            }
        }
    },
    {
        $match: {
            $expr: {
                $gte: [{
                    $size: "$score"
                }, 5]
            }
        }
    }
])

使用最终的 $match 和 $expr 来执行聚合表达式