Question

是否可以使用聚合框架按数组的特定元素进行分组？

使用这样的文件：

{
  name: 'Russell',
  favourite_foods: [
    { name: 'Pizza', type: 'Four Cheeses' },
    { name: 'Burger', type: 'Veggie'}
  ],
  height: 6
}

我可以得到一份最受欢迎的食物清单（即0号食物）以及最受欢迎食物的最高人物的高度？

像这样的东西（虽然它不起作用，因为数组索引访问点表示法似乎在聚合框架中不起作用）：

db.people.aggregate([
  { $group : { _id: "$favourite_foods.0.name", max_height: { $max : "$height" } } }
])

Answer 1

好像你依赖于每个人在阵列中的第一个最喜欢的食物。如果是这样，您可以利用聚合框架运算符。

以下是您可以使用的管道：

db.people.aggregate(
[
    {
        "$unwind" : "$favourite_foods"
    },
    {
        "$group" : {
            "_id" : {
                "name" : "$name",
                "height" : "$height"
            },
            "faveFood" : {
                "$first" : "$favourite_foods"
            }
        }
    },
    {
        "$group" : {
            "_id" : "$faveFood.name",
            "height" : {
                "$max" : "$_id.height"
            }
        }
    }
])

在此样本数据集上：

> db.people.find().pretty()
{
    "_id" : ObjectId("508894efd4197aa2b9490741"),
    "name" : "Russell",
    "favourite_foods" : [
        {
            "name" : "Pizza",
            "type" : "Four Cheeses"
        },
        {
            "name" : "Burger",
            "type" : "Veggie"
        }
    ],
    "height" : 6
}
{
    "_id" : ObjectId("5088950bd4197aa2b9490742"),
    "name" : "Lucy",
    "favourite_foods" : [
        {
            "name" : "Pasta",
            "type" : "Four Cheeses"
        },
        {
            "name" : "Burger",
            "type" : "Veggie"
        }
    ],
    "height" : 5.5
}
{
    "_id" : ObjectId("5088951dd4197aa2b9490743"),
    "name" : "Landy",
    "favourite_foods" : [
        {
            "name" : "Pizza",
            "type" : "Four Cheeses"
        },
        {
            "name" : "Pizza",
            "type" : "Veggie"
        }
    ],
    "height" : 5
}
{
    "_id" : ObjectId("50889541d4197aa2b9490744"),
    "name" : "Augie",
    "favourite_foods" : [
        {
            "name" : "Sushi",
            "type" : "Four Cheeses"
        },
        {
            "name" : "Pizza",
            "type" : "Veggie"
        }
    ],
    "height" : 6.2
}

你得到这些结果：

{
    "result" : [
        {
            "_id" : "Pasta",
            "height" : 5.5
        },
        {
            "_id" : "Pizza",
            "height" : 6
        },
        {
            "_id" : "Sushi",
            "height" : 6.2
        }
    ],
    "ok" : 1
}

Answer 2

看起来目前无法在聚合中从数组中提取特定元素： https://jira.mongodb.org/browse/SERVER-4589

Answer 3

我认为你可以使用$ project和$ unwind运算符（让我知道这不是你想要完成的）：

> db.people.aggregate(
   {$unwind: "$favourite_foods"}, 
   {$project: {food : "$favourite_foods", height: 1}}, 
   {$group : { _id: "$food", max_height: { $max : "$height" } } })


{
    "result" : [
        {
            "_id" : {
                "name" : "Burger",
                "type" : "Veggie"
            },
            "max_height" : 6
        },
        {
            "_id" : {
                "name" : "Pizza",
                "type" : "Four Cheeses"
            },
            "max_height" : 6
        }
    ],
    "ok" : 1
}

http://docs.mongodb.org/manual/applications/aggregation/

Answer 4

在使用"$wind"后，我必须添加有关结果的更多信息：

文件：

> db.people.find().pretty()
{
    "_id" : ObjectId("508894efd4197aa2b9490741"),
    "name" : "Russell",
    "favourite_foods" : [
        {
            "name" : "Pizza",
            "type" : "Four Cheeses"
        },
        {
            "name" : "Burger",
            "type" : "Veggie"
        }
    ],
    "height" : 6
},
...

AGGREAGATION：

db.people.aggregate([{
    $unwind: "$favourite_foods"
}]);

结果：

{
    "_id" : ObjectId("508894efd4197aa2b9490741"),
    "name" : "Russell",
    "favourite_foods" :{
            "name" : "Pizza",
            "type" : "Four Cheeses"
    },
    "height" : 6
},
{
    "_id" : ObjectId("508894efd4197aa2b9490741"),
    "name" : "Russell",
    "favourite_foods" : {
            "name" : "Burger",
            "type" : "Veggie"
    },
    "height" : 6
}

另外：
如果一个集合记录中有两个以上的数组字段，我们可以使用"$project" stage来指定数组字段。

db.people.aggregate([
    {
        $project:{
            "favourite_foods": 1
        }
    },
    {
        $unwind: "$favourite_foods"
    }
]);

使用mongo聚合框架按数组的特定元素进行分组

4 个答案: