MongoDB的：带2个嵌套数组更新文档中的平均 [英] MongoDB: Updating an average in a document with 2 nested arrays

问题描述

我有以下MongoDB的文档：

I have the following MongoDB document:

{
    _id: ObjectId(),
    company_name: "Name",
    registered: 2/21/2015 2:00,
    trucks: [
        {
            truck_id: "TEB7622",
            weight: 88.33,
            capacity: 273.333,
            length: 378.333,
            width: 377.383,
            average_grade: 2.5,
            grades: [
                {
                    grade_number: 4,
                    timestamp: 2/21/2015 2:00
                }
            ]
        },
        {
            truck_id: "TEB5572",
            weight: 854.33,
            capacity: 2735.333,
            length: 378.333,
            width: 37.383,
            average_grade: 3.8,
            grades: [
                {
                    grade_number: 4,
                    timestamp: 2/21/2015 2:00
                }
            ]
        }
    ]

}

我希望每个卡车的 average_grade 加入所有的 grade_numbers 更新。我遇到的问题是， grade_numbers 我想加在一个阵列内的阵列。我一直在使用 $放松来放松两个卡车和档次阵列尝试。

I want to update each truck's average_grade by adding all the grade_numbers. The problem I'm having is that the grade_numbers I am trying to add are in an array within an array. I have tried using $unwind to unwind both trucks and grades arrays.

这是我已经尝试使用查询：

This is the query I have tried using:

db.col.aggregate([ 
   {$unwind: "$trucks"}, 
   {$unwind: "$trucks.grades"}, 
   { $project: { 
      "_id": "$trucks.truck_id", 
      "trucks.average_grade": { $avg: { $sum: "trucks.grades.grade_number"} } 
      } 
   }])

我需要更多的东西添加到我的查询？我要更新所有的 trucks.average_grades ，因为我试图更新文档中的很多人。

Do I have to add something more to my query? I want to update ALL of the trucks.average_grades, since there a lot of them in the document I am trying to update.

推荐答案

您不能使用汇聚更新文件，但你绝对可以用它来得到你的数据要使用更新。首先，我注意到有一些 {} 在你的品位缺少的等级阵列。您可能要仔细检查你的文档结构作为发布。其次，与您的聚集查询几个问题。

You cannot use aggregation to update a document, but you can definitely use it to get the data you want to use for an update. First of all, I noticed that there are some {} missing around your grade object inside the grades array. You might want to double check that your document structure is as posted. Secondly, there are a couple of issues with your aggregation query.

1. 的 $平均运营商工作的 $组子句中，而不是一个 $项目。


2. 当您使用 $平均，你并不需要使用 $金额。


3. 您要平均 trucks.grades.grade.grade_number ，其实际持有的档次数值。也就是说，你缺少等级之间的等级和 GRADE_NUMBER

1. The $avg operator works inside a $group clause, not a $project.
2. When you use $avg, you do not need to use $sum.
3. You want to average trucks.grades.grade.grade_number, which actually holds the numeric value of the grade. That is, you are missing grade between grades and grade_number.

如果你解决这些问题，你会得到类似以下的查询：

If you resolve those issues, you get a query similar to the following:

db.col.aggregate([ 
    { "$unwind": "$trucks" }, 
    { "$unwind": "$trucks.grades" }, 
    { "$group":
        { 
            "_id": "$trucks.truck_id", 
            "average_grade": { "$avg": "$trucks.grades.grade_number" } 
        } 
    }
]);

有关示例文档，返回：

{ "_id" : "TEB5572", "average_grade" : 4 }
{ "_id" : "TEB7622", "average_grade" : 4 }

现在你可以使用这个信息来更新 average_grade 字段。如果您使用的MongoDB版本2.6或更高时，总方法将返回一个指针。您可以通过迭代游标并相应地更新文档。

Now you can use this information to update the average_grade field. If you're using MongoDB version 2.6 or higher, the aggregate method will return a cursor. You can iterate through that cursor and update the documents accordingly.

在这个例子中，我搜索具有特定 truck_id 的卡车内阵列，并继续更新文件在 average_grade 通过一个又聚集查询的计算。您可以修改它来满足您的需求。与聚集查询相结合，code看起来是这样的。

In this example, I search for documents that have a particular truck_id inside their trucks array and proceed to update the average_grade with the one computed by the aggregation query. You can modify it to suit your needs. Combined with the aggregation query, the code looks like this.

// Get average grade for each truck and assign results to cursor.
var cur = db.col.aggregate([ 
    { "$unwind": "$trucks" }, 
    { "$unwind": "$trucks.grades" }, 
    { "$group":
        { 
            "_id": "$trucks.truck_id", 
            "average_grade": { "$avg": "$trucks.grades.grade_number" } 
        } 
    }
]);

// Iterate through results and update average grade for each truck.
while (cur.hasNext()) {
    var doc = cur.next();
    db.col.update({ "trucks.truck_id": doc._id },
                  { "$set": { "trucks.$.average_grade": doc.average_grade }},
                  { "multi": true});
}

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