结合$First和$last结果的Mongo

Question

我有这样的数据集

{ "_id" : { "borough" : "Manhattan", "cuisine" : "Tex-Mex" }, "RestaruntCount" : 53 }
{ "_id" : { "borough" : "Manhattan", "cuisine" : "Bakery" }, "RestaruntCount" : 221 }
{ "_id" : { "borough" : "Manhattan", "cuisine" : "Soups & Sandwiches" }, "RestaruntCount" : 44 }
{ "_id" : { "borough" : "Manhattan", "cuisine" : "Vietnamese/Cambodian/Malaysia" }, "RestaruntCount" : 38 }
{ "_id" : { "borough" : "Manhattan", "cuisine" : "Filipino" }, "RestaruntCount" : 5 }
{ "_id" : { "borough" : "Manhattan", "cuisine" : "Egyptian" }, "RestaruntCount" : 5 }
{ "_id" : { "borough" : "Manhattan", "cuisine" : "Nuts/Confectionary" }, "RestaruntCount" : 4 }

从这个数据集，我需要得到最高和最低的美食类型的数量在曼哈顿区。我知道我需要使用$first和$last来实现它，但是我无法继续执行它。

目前，我从下面提到的查询中获得了这个输出：

db.restaurants.aggregate(    [      { $match: { "borough": "Manhattan"  } },     { $group: {_id: {borough:"$borough", cuisine: "$cuisine"}, count : { $sum : 1} } },  {$sort: {count:1,_id:1}}     ] );

Answer 1

尝试以下聚合：

db.restaurants.aggregate([
    {
        $match: { "_id.borough": "Manhattan" }
    },
    {
        $sort: { RestaruntCount: 1 }
    },
    {
        $group: {
            _id: {borough:"$borough", cuisine: "$cuisine"},
            first: { $first: "$$ROOT" },
            last: { $last: "$$ROOT" }
        }
    }
])

您可以使用特殊变量$$ROOT在聚合期间捕获整个文档。