MongoDB通过聚合分组

如何解决MongoDB通过聚合分组

我有一个包含物品的集合，我想要的是一个汇总，其中我可以将所有item.recipes.life_skill.names按加热，烹饪等名称进行分组

我试图建立汇总，但我仍然不知道

{
  _id: "recipes.life_skill.name","life_skill": {
    $first: "recipes.life_skill.name"
  }
}

以下是此类物品的示例：

{
   "_id": {
      "$oid": "5f87ddc65a997f69083d663d"
   },"id": "756005","global_id": "item--756005","category": "Special Items","name": "Lunar Black Stone","grade": 3,"weight": "0.01","required_level": 1,"url": null,"icon_url": null,"recipes": [{
       "global_id": "mrecipe--2103","id": "2103","life_skill": {
           "name": "Heating","rank": "Beginner","level": 1
       },"xp_amount": null,"input_products": [{
           "type": "ITEM","id": "756002","name": "Frosted Black Stone","quantity": 10,"url": null
       },{
           "type": "ITEM","id": "756003","name": "Starlight Crystal","quantity": 1,"url": null
       }],"output_products": [{
           "type": "ITEM","product_groups": []
   }],"used_in_recipes": null,"ship_upgrades": null,"achievements": null
}

解决方法

您需要先执行$unwind来解构数组。

db.collection.aggregate([
  {
    $unwind: "$recipes"
  },{
    $group: {
      _id: {
        _id: "$_id",life_skill: "$recipes.life_skill.name"
      },life_skill: {
        $first: "$recipes.life_skill.name"
      }
    }
  }
])

由于您只给出了一个文档，所以我使用组合_id进行分组

_id: {
        _id: "$_id",life_skill: "$recipes.life_skill.name"
      }

这有助于按对象明智地分组。如果您需要对所有对象进行整体分组，只需使用

_id: "$recipes.life_skill.name"

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