为什么 pymongo MongoDB 在 find_one() 中不返回精确值？

发布于 2025-01-11 10:27:59 字数 953 浏览 1 评论 0原文

我想从 pymongo DB 检索单个值“count”，但它不起作用。下图显示了如何设置数据输入。

这是对我的数据库类的调用以使用 db.find_one()。

这里的代码：

    filters = {"email": session.get('email')}
    returns = {f'words.{today_s}.{self.length - 3}.count': 1}

    count_value = Database.find_one_return_one("users", filters, returns)

    print({f'words.{today_s}.{self.length - 3}.count':1})
    print(count_value)

@staticmethod
def find_one_return_one(collection: str, query: Dict, data: Dict) -> Dict:
    return Database.DATABASE[collection].find_one(query, data)

这会从正确的数据返回一个空的字典列表？我想要返回计数值。

这是投影查询： {words.20220302.0.count : 1}

这是返回的内容：

{'_id': ObjectId('621ee5065d08c44070140df0'), 'words': {'20220302': [{}, {}, {}, {}, {}, {}, {}]}}

出了什么问题或者是否有更好更快的方法来检索计数值？

原文

I want to retrieve the single value "count "from pymongo DB but it is not working. The image below shows how the data entry is setup.

Here is the call to my Database class to use the db.find_one().

CODE HERE:

    filters = {"email": session.get('email')}
    returns = {f'words.{today_s}.{self.length - 3}.count': 1}

    count_value = Database.find_one_return_one("users", filters, returns)

    print({f'words.{today_s}.{self.length - 3}.count':1})
    print(count_value)

@staticmethod
def find_one_return_one(collection: str, query: Dict, data: Dict) -> Dict:
    return Database.DATABASE[collection].find_one(query, data)

This returns an empty list of dictionaries from the correct data? I want the count value returned.

This is the projection query: {words.20220302.0.count : 1}

This is what is returned:

{'_id': ObjectId('621ee5065d08c44070140df0'), 'words': {'20220302': [{}, {}, {}, {}, {}, {}, {}]}}

What is wrong or is there a better quicker way to retrieve the count value?

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つ可否回来 2025-01-18 10:27:59

可以使用以下查询投影来获得所需的结果。请注意，这适用于 MongoDB v5。

样本文件；与问题帖子中的类似：

{ _id: 1, words: { fld: [ { a: 1, b: 2 }, { a: 9, b: 100 } ] } }

预期结果是： { "_id" : 1, "words" : { "fld" : { "a" : 9 } } }

查询：

INDEX = 1    # this is the index of the array element
query = { }
projection = { 
    'words.fld': { 
        '$arrayElemAt': [ 
            { '$map': { 'input': '$words.fld', 'in': { 'a': '$this.a' } } }, 
            INDEX 
        ] 
    }
}

result = collection.find_one(query, projection)
print(result)

The following query projection can be used to get the desired result. Note this worked with MongoDB v5.

A sample document; similar to the one in the question post:

{ _id: 1, words: { fld: [ { a: 1, b: 2 }, { a: 9, b: 100 } ] } }

The expected result is: { "_id" : 1, "words" : { "fld" : { "a" : 9 } } }

The query:

INDEX = 1    # this is the index of the array element
query = { }
projection = { 
    'words.fld': { 
        '$arrayElemAt': [ 
            { '$map': { 'input': '$words.fld', 'in': { 'a': '$this.a' } } }, 
            INDEX 
        ] 
    }
}

result = collection.find_one(query, projection)
print(result)

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