如何为上述XML标记创建模型类？ [英] How to create model class for the mentioned XML tag ?

问题描述

如何为xml标记创建模型类。在反序列化获取异常时，模型类的reflcetion类型中存在错误。

我的Xml如下：

How to create model class for the xml tag. On deserializing getting the exception there is an error in reflcetion type of Model class.
My Xml is as below:

<advice_tags>
  
  <aclassname>SAV1 ACCOUNT</aclassname>
  <blockedbalance>0</blockedbalance>
  <acy>
    <blocked_acy>23</blocked_acy>
    <amount_acy>56</amount_acy>
  </acy>
  
</advice_tags>

我上面的Xml标签的Model类如下：

My Model class for the above Xml Tag is as below:

[XmlElement("ACLASSNAME")]
       public string ACLASSNAME { get; set; }
       [XmlElement("BLOCKEDBALANCE")]
       public string BLOCKEDBALANCE { get; set; }


       [XmlArray("ACY")]
       [XmlArrayItem("BLOCKED_ACY")]
       [XmlArrayItem("AMOUNT_ACY")]
       public List<string> ACY { get; set; }

但是我对XmlArray属性进行脱盐处理时遇到异常。

But I am getting exception on desalinizing the same for the XmlArray attribute.

MDL_UBSFile UBSFile = null;
                                  XmlSerializer serializer = new XmlSerializer(typeof(MDL_UBSFile));
                                  reader = new StreamReader(Path.Combine(UBSFilePath, FileName));
                                  dss.ReadXml(reader);
                                  reader.Close();

推荐答案

你的xml对大小写字母都很敏感，所以说

Your xml become sensitive to big and small letters, so say

<myclass>
   <myproperty>1</myproperty>
</myclass>

不是

is not

public class MyClass{
  public int MyProperty{ get;set;}
}

宁愿是

rather would be

public class myclass{
  public int myproperty{ get;set;}
}

所以你需要做的就是匹配你的写作方式



......（更新解决方案）......

所以如果这不起作用，我建议走另一条路。让课程首先



通过观察你的xml很清楚，可以使用两个简单类型和一个名为Acy的复杂类型生成一个类AdviceTag，其中包含两个值（Blocked）和金额），或许是这样的：

so what you need to do is match the way you wrote it

... (updating solution) ...
So if that doesn't work, i suggest going the other way about. Make the class first

and it is clear by observing your xml that a class AdviceTag could be generated with two simply types and one complex type called Acy with two values (Blocked and amount), perhaps like this:

public class AdviceTag
    {
        public string AClassName { get; set; }
        public string BlockedBalance { get; set; }

        public Acy Acy { get; set; }
    }

    public class Acy
    {
        public string Blocked{ get; set; }
        public string Amount { get; set; }
    }

这样就可以通过实例化该类的实例并将其序列化来生成xml。 />

so that in place let's simply generate the xml by instantiating an instance of that class and serialize it

var aTag = new AdviceTag { AClassName = "Sav1 Account", BlockedBalance = "0", Acy = new Acy { Amount = "56", Blocked = "23" } };

           var ser = new XmlSerializer(typeof(AdviceTag));

           string filename = Assembly.GetExecutingAssembly().Location;
           int pos = filename.LastIndexOf(@"\");
           filename = filename.Substring(0, pos +1) + "myfile.xml";
           using(var fs = new FileStream(filename, FileMode.Create, FileAccess.Write))
           {
               ser.Serialize(fs, aTag);
               fs.Flush();
               fs.Close();
           }

我们将序列化的xml放在与执行程序集相同的位置，并将其命名为myfile.xml，然后看起来像这样：

We're placing the serialized xml in same location as executing assembly and calling it myfile.xml, it then looks like this:

<?xml version="1.0"?>
<AdviceTag xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <AClassName>Sav1 Account</AClassName>
  <BlockedBalance>0</BlockedBalance>
  <Acy>
    <Blocked>23</Blocked>
    <Amount>56</Amount>
  </Acy>
</AdviceTag>

现在出现了什么错误？



那么你的xml指的是一个复杂的类型名为acy，包含两个成员属性，但你很难将它们放在一个arraylist中，你不必这样做。

And now what was the mistake?

Well your xml designates a complex type called acy containing two member properties, but you were struggleing to put them in an arraylist, where you don't have to.

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