我有一个包含项的集合，我想要的是一个聚合，其中我得到了所有item.recipes.life_skill.names按名称进行分组，如加热，烹饪等
我试图建立聚合，但我仍然不知道
{
_id: "recipes.life_skill.name",
"life_skill": {
$first: "recipes.life_skill.name"
}
}
这是此类物品的示例：
{
"_id": {
"$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,
"url": 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",
"id": "756005",
"name": "Lunar Black Stone",
"quantity": 1,
"url": null
}],
"product_groups": []
}],
"used_in_recipes": null,
"ship_upgrades": null,
"achievements": null
}
您需要首先执行$unwind以解构阵列。
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，所以我习惯于_id组合
_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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