Question

假设我们有以下来自聚合管道的文档：

[
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "description": "description for item 1",
    "item_code": "00001"
  },
  {
    "_id": ObjectId("5a934e000102030405000001"),
    "description": "description for item 2",
    "item_code": "00002"
  },
  {
    "_id": ObjectId("5a934e000102030405000002"),
    "description": "description for item 3",
    "item_code": "00003"
  },
  {
    "_id": ObjectId("5a934e000102030405000003"),
    "extrafield": "extra field for item 2",
    "item_code": "00002"
  }
]

如何将具有相同item_code的文档合并为一个文档，同时保留所有属性？所需结果：

[
  {
    "description": "description for item 1",
    "item_code": "00001"
  },
  {
    "description": "description for item 2",
    "extrafield": "extra field for item 2",
    "item_code": "00002"
  },
  {
    "description": "description for item 3",
    "item_code": "00003"
  }
]

我尝试了不同的$group模式，但没有成功：（

Here's mongodb playground

Answer 1

您可以尝试

$group by item_code，将$mergeObjects与$$ROOT合并对象
$replaceWith将根对象替换为根

db.collection.aggregate([
  {
    $group: {
      _id: "$item_code",
      root: { $mergeObjects: "$$ROOT" }
    }
  },
  { $replaceWith: "$root" }
])

Playground

mongodb-根据条件字段合并两个对象

1 个答案: