Question

我有一个具有以下结构的集合：

{
        "_id" : ObjectId("59ef54445134d7d70e1cf531"),
        "CustomerId" : "Gym_2",
        "History" : [
                {
                        "Created_At" : ISODate("2017-10-24T14:54:59Z"),
                        "Unit" : 600,
                        "ReferenceCode" : "1cd15b4d-bc42-4a51-a8b3-307db6dc3dee",

                },
                {
                        "Created_At" : ISODate("2017-10-28T00:22:19Z"),
                        "Sent" : true
                },
                {
                        "Created_At" : ISODate("2017-10-29T10:22:23Z"),
                        "Unit" : 600,
                        "ReferenceCode" : "998e7fce-8a1c-4f7c-b48c-c02cb5c5ad5c",
                }
        ]
}
{
        "_id" : ObjectId("59ef54465134d7d70e1cf534"),
        "CustomerId" : "Gym_1",
        "History" : [
                {
                        "Created_At" : ISODate("2017-10-24T14:55:02Z"),
                        "Unit" : 600,
                        "ReferenceCode" : "d19ebeec-bd81-4a0a-aed5-006f746b50ff",
                },
                {
                        "Unit" : 600,
                        "ReferenceCode" : "a991504f-be1f-4e77-b59f-fba73c59e6f1",
                        "Created_At" : ISODate("2017-10-26T13:51:14Z")
                }
        ]
}

我尝试构建一个只返回CustomerId的查询以及没有设置"Sent"字段的历史对象。结果应如下所示：

       {
                "_id" : ObjectId("59ef54445134d7d70e1cf531"),
                "CustomerId" : "Gym_2",
                "History" : [
                        {
                                "Created_At" : ISODate("2017-10-24T14:54:59Z"),
                                "Unit" : 600,
                                "ReferenceCode" : "1cd15b4d-bc42-4a51-a8b3-307db6dc3dee",

                        },

    {
                            "Created_At" : ISODate("2017-10-29T10:22:23Z"),
                            "Unit" : 600,
                            "ReferenceCode" : "998e7fce-8a1c-4f7c-b48c-c02cb5c5ad5c",
                    }
            ]
    }
    {
            "_id" : ObjectId("59ef54465134d7d70e1cf534"),
            "CustomerId" : "Gym_1",
            "History" : [
                    {
                            "Created_At" : ISODate("2017-10-24T14:55:02Z"),
                            "Unit" : 600,
                            "ReferenceCode" : "d19ebeec-bd81-4a0a-aed5-006f746b50ff",
                    },
                    {
                            "Unit" : 600,
                            "ReferenceCode" : "a991504f-be1f-4e77-b59f-fba73c59e6f1",
                            "Created_At" : ISODate("2017-10-26T13:51:14Z")
                    }
            ]
    }

我能够达到的最接近的是以下查询：

db.Customers.aggregate([ 
  {$project:{"Sent":{$exists:false},count:{$size:"$History" }}}
  ]);

但我得"errmsg" : "Unrecognized expression '$exists'"。我怎样才能达到这个结果？

Answer 1

您的问题有一个解决方案，聚合框架绝对是实现您想要的正确方法。要修改嵌套集合，您必须：

展开此系列（$ unwind）
过滤掉已存在的文档
按共同属性分组

投放您的数据以接收原始表格

import java.io.FileWriter
class Main{

    static void main(String[] args)
    {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in))
        println "Enter the number of items:"
        int n = Integer.parseInt(br.readLine())
        String[] item =new String[n]
        Double[] price =new Double[n]
        int[] quantity = new int[n]

        for(int i=0; i<n;i++)
        {
            int it = i+1
            println "Enter item $it details: "
            item[i] = br.readLine()
            price[i] = Double.parseDouble(br.readLine()).round(2)
            quantity[i] = Integer.parseInt(br.readLine())
        }

        String COMMA_DELIMITER = ","
        String NEW_LINE_SEPARATOR = "\n"

        try
        {
            FileWriter fileWriter = new FileWriter("invoicedetails.csv")

            for(int j=0; j<n;j++)
            {
                fileWriter.append(item[j])
                fileWriter.append(COMMA_DELIMITER)
                fileWriter.append(price[j])
                fileWriter.append(COMMA_DELIMITER)
                fileWriter.append(quantity[j])
                fileWriter.append(NEW_LINE_SEPARATOR)   
            }
        }
        catch(Exception e)
        {
            println e
        } 
    }
}

正如您所看到的，此查询相当复杂，对于较大的集合，您可能会遇到性能问题，因为我们做的不仅仅是简单的集合过滤。因此，虽然它有效但我建议您应该考虑更改数据模型，例如将每个历史项目作为单独的文档，如下所示：

db.customers.aggregate([
   {$unwind: "$History"},
   {$match: {"History.Sent": {$exists: false}}},
   {$group: {"_id": { "_id": "$_id", "CustomerId": "$CustomerId" }, History: { $push: "$History"} }},
   {$project: { "_id": "$_id._id", "CustomerId": "$_id.CustomerId", History: 1}}
]);

然后您的查询将只是简单查找$ exists。

如何在mongodb中从数组中排除特定对象？

1 个答案: