MongoDB |排序不适用于聚合

Question

我需要使用聚合来获取不同的值，并且需要按字符串对它们进行排序。在这种情况下，我有一个集合，其中包含很多关于地点的文档。我需要编写一个代码，输出位于“布朗克斯”的前十个地点，按地点名称排序（从 z 到 a）- 并且只显示地点名称。

JSON 中一份文件的样本。有许多相似但具有不同值的对象：

{
  "borough": "Bronx",
  "cuisine": "Bakery",
  "name": "Champion Bakery",
  "restaurant_id": "40423830"
}

borough:Bronx 有多个重复名称，因此我需要使用聚合来仅获取不同的值。

我尝试使用不同的命令，但没有任何效果 - 我确实得到了不同的值，但它们往往是随机的并且没有排序。使用查找，排序确实有效。我在这里检查了各种答案，但没有任何帮助。

我希望输出的排序方式与此类似，但没有重复值：

{ "name" : "Zaro'S Bread Basket" }
{ "name" : "Yolanda Pizzeria Restaurant" }
{ "name" : "Yankee Tavern" } 
{ "name" : "Yankee Jz Pizza" } 
{ "name" : "Woodlawn Cafe" } 
{ "name" : "Wild Asia" } 
{ "name" : "Wilbel Pizza" } 
{ "name" : "White Castle" } 
{ "name" : "White Castle" }
{ "name" : "White Castle" }

我试过的代码 + 输出：

db.places.aggregate([
  { $match: { borough: "Bronx" } },
  { $group: { _id: "$name" } },
  { $sort: { name: -1 } },
  { $limit: 10 }
]).pretty()

输出：

{ "_id" : "Morris Park Bake Shop" }
{ "_id" : "Roca Tone Seafood Restaurant" }
{ "_id" : "Rolen Bagels" }
{ "_id" : "Casa Promesa" }
{ "_id" : "Celeste'S Snack Bar" }
{ "_id" : "The Pizza Place" }
{ "_id" : "Bronx Grill" }
{ "_id" : "Pepe Joes Pizzeria" }
{ "_id" : "Ray'S Pizza Restuarant" }
{ "_id" : "Boulevard Tavern" }

db.places.aggregate([
  { $match: { borough: "Bronx" } },
  { $sort: { name: -1 } },
  { $group: { _id: "$name" } },
  { $limit: 10 }
]).pretty()

输出：

{ "_id" : "Pepe Joes Pizzeria" }
{ "_id" : "Bronx Grill" }
{ "_id" : "Ray'S Pizza Restuarant" }
{ "_id" : "The Pizza Place" }
{ "_id" : "Rolen Bagels" }
{ "_id" : "Celeste'S Snack Bar" }
{ "_id" : "Casa Promesa" }
{ "_id" : "Boulevard Tavern" }
{ "_id" : "Morris Park Bake Shop" }
{ "_id" : "Roca Tone Seafood Restaurant" }

db.places.aggregate([
  { $match: { borough: "Bronx" } },
  { $sort: { name: -1 } },
  { $group: { _id: "$name" } },
  { $sort: { name: -1 } },
  { $limit: 10 }
]).pretty()

输出：

{ "_id" : "Morris Park Bake Shop" }
{ "_id" : "Roca Tone Seafood Restaurant" }
{ "_id" : "Rolen Bagels" }
{ "_id" : "Boulevard Tavern" }
{ "_id" : "Celeste'S Snack Bar" }
{ "_id" : "The Pizza Place" }
{ "_id" : "Bronx Grill" }
{ "_id" : "Pepe Joes Pizzeria" }
{ "_id" : "Ray'S Pizza Restuarant" }
{ "_id" : "Casa Promesa" }

提前致谢！

Answer 1

您的代码运行良好。如您所见，您 { $sort: { _id: -1 } 然后字母 T 将在顶部，字母 B 在最后。

db.collection.aggregate([
  {
    $match: {
      borough: "Bronx"
    }
  },
  {
    $group: {
      _id: "$name"
    }
  },
  {
    $sort: {
      _id: -1
    }
  },
  {
    $limit: 10
  },
  {
    $project: {
      _id: 0,
      name: "$_id"
    }
  }
])

输出

[
  {
    "name": "Zaro'S Bread Basket"
  },
  {
    "name": "Yolanda Pizzeria Restaurant"
  },
  {
    "name": "Yankee Tavern"
  },
  {
    "name": "Yankee Jz Pizza"
  },
  {
    "name": "Woodlawn Cafe"
  },
  {
    "name": "Wild Asia"
  },
  {
    "name": "Wilbel Pizza"
  },
  {
    "name": "White Castle"
  },
  {
    "name": "The Pizza Place"
  },
  {
    "name": "Rolen Bagels"
  }
]

mongoplayground

MongoDB |排序不适用于聚合

MongoDB | sort doesn't work with aggregate

sorting

aggregate

mongodb