来自嵌套数组 MongoDB 的项目对象

发布于 2025-01-13 23:15:03 字数 513 浏览 0 评论 0原文

[
  {
    "_id": "grandParentId",
    "types": [
      {
        "_id": "parentId",
        "files": [
          {
            "url": "1.example.com",
            "_id": "1childId"
          },
          {
            "url": "2.example.com",
            "_id": "2childId"
          }
        ]
      }
    ]
  }
]

cond: { _id: 2childId }

预期输出：

{ 
    "url": "2.example.com",
    "_id": "2childId"
}

问题 2：这是一个好方法还是我应该使用循环来获得所需的输出？

原文

[
  {
    "_id": "grandParentId",
    "types": [
      {
        "_id": "parentId",
        "files": [
          {
            "url": "1.example.com",
            "_id": "1childId"
          },
          {
            "url": "2.example.com",
            "_id": "2childId"
          }
        ]
      }
    ]
  }
]

cond: { _id: 2childId }

expected output:

{ 
    "url": "2.example.com",
    "_id": "2childId"
}

question 2: is this a good approach or should I just use loops to get the desired output?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

云雾 2025-01-20 23:15:03

使用聚合 $unwind 和 $project

在 mongoPlayground< 进行测试/a>

db.collection.aggregate([
  {
    "$unwind": "$types"
  },
  {
    "$unwind": "$types.files"
  },
  {
    "$match": {
      "types.files._id": "2childId"
    }
  },
  {
    "$project": {
      "url": "$types.files.url",
      "_id": "$types.files._id"
    }
  }
])

useing aggregate $unwind and $project

test it at mongoPlayground

db.collection.aggregate([
  {
    "$unwind": "$types"
  },
  {
    "$unwind": "$types.files"
  },
  {
    "$match": {
      "types.files._id": "2childId"
    }
  },
  {
    "$project": {
      "url": "$types.files.url",
      "_id": "$types.files._id"
    }
  }
])

回复收藏 0 原文

面如桃花 2025-01-20 23:15:03

我确信还有其他方法，但您可以使用 mongo 聚合框架：

db.collection.aggregate([
  {
    $match: {
      "types.files._id": "2childId"
    }
  },
  {
    $unwind: "$types"
  },
  {
    $unwind: "$types.files"
  },
  {
    $replaceRoot: {
      newRoot: "$types.files"
    }
  },
  {
    $match: {
      _id: "2childId"
    }
  }
])

第一个 $match 只是过滤与您的条件匹配的文档。 $unwind 使用两次来解构不同的数组。 $replaceRoot 将输入文档替换为指定文档（在本例中为子文档，files）和最终的$match 过滤器 files 我们之前根据您的情况解构的数组。第一个 $match 只是出于性能原因，您可以根据需要删除它。

来了解阶段的工作原理

您可以在 https://mongoplayground.net/p/hoz24xP_BNV查看 mongo 聚合框架如果您想了解更多信息，请参阅文档（https://docs.mongodb.com/manual/aggregation/）

I'm sure that there are other ways, but you can use mongo aggregation framework:

db.collection.aggregate([
  {
    $match: {
      "types.files._id": "2childId"
    }
  },
  {
    $unwind: "$types"
  },
  {
    $unwind: "$types.files"
  },
  {
    $replaceRoot: {
      newRoot: "$types.files"
    }
  },
  {
    $match: {
      _id: "2childId"
    }
  }
])

The first $match is just to filter documents which match with your condition. $unwind are used two times to deconstruct the different arrays. $replaceRoot replaces the input document with the specified document (in this case subdocument, files) and final $match filters files arrays that we deconstructed previously, based on your condition. First $match is just for performance reasons, you can delete it if you want.

You can see how stages works in https://mongoplayground.net/p/hoz24xP_BNV

See mongo aggregation framework documentation if you want to learn more about it (https://docs.mongodb.com/manual/aggregation/)

回复收藏 0 原文

~没有更多了~