如何根据MongoDB中对象中的值对记录进行排序

发布于 2025-01-24 17:24:35 字数 880 浏览 0 评论 0原文

请考虑以下MongoDB集合

[
    {
        _id: 123123,
        name: "abc"
        topic: {
            asda: "Z"
        }
    },
    {
        _id: 123123,
        name: "abc"
        topic: {
            dasd: "P"
        }
    },{
        _id: 123123,
        name: "abc"
        topic: {
            qwer: "A"
        }
    }
]

根据主题对象中的值进行了查询需求记录（casemensitiationality casemensitiationality ossensitiationality offictial otiven opivent otiven对象）之后，

。注意：主题对象内的密钥将始终是每个记录的不同唯一字符串，

因此结果记录应如下所示

[
    {
        _id: 123123,
        name: "abc"
        topic: {
            qwer: "A"
        }
    },
    {
        _id: 123123,
        name: "abc"
        topic: {
            dasd: "P"
        }
    },{
        _id: 123123,
        name: "abc"
        topic: {
            qwer: "Z"
        }
    }
]

：:)

原文

Consider the following MongoDB Collection

[
    {
        _id: 123123,
        name: "abc"
        topic: {
            asda: "Z"
        }
    },
    {
        _id: 123123,
        name: "abc"
        topic: {
            dasd: "P"
        }
    },{
        _id: 123123,
        name: "abc"
        topic: {
            qwer: "A"
        }
    }
]

After performing a query need records sorted(case-insensitively) based on the value inside the topic object.

Note: The key inside the topic object will be always a different and unique string for every record

So the result records should be as follows

[
    {
        _id: 123123,
        name: "abc"
        topic: {
            qwer: "A"
        }
    },
    {
        _id: 123123,
        name: "abc"
        topic: {
            dasd: "P"
        }
    },{
        _id: 123123,
        name: "abc"
        topic: {
            qwer: "Z"
        }
    }
]

Thanks in Advance :)

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枫林﹌晚霞¤ 2025-01-31 17:24:35

如果要以未知键的值排序结果，则可以使用$ objectToArray

db.collection.aggregate([
  {
    $addFields: {
      topicArr: {
        $objectToArray: "$topic"
      }
    }
  },
  {
    $sort: {
      "topicArr.v": 1
    }
  },
  {
    $unset: "topicArr"
  }
])

在此游乐场示例

If you want to get the results sorted by the value of an unknown key, you can use $objectToArray

db.collection.aggregate([
  {
    $addFields: {
      topicArr: {
        $objectToArray: "$topic"
      }
    }
  },
  {
    $sort: {
      "topicArr.v": 1
    }
  },
  {
    $unset: "topicArr"
  }
])

As you can see on this playground example

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