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在尝试让我的编程反序列化多个 XmlArrayItem 时，我遇到了一个相当愚蠢的问题。

XML 基本上如下所示：

<Root2>
   <Data2>
     <HOLD>
      ...
     </HOLD>
     <CUST_HOLD>
      ...
     </CUST_HOLD>
   </DATA2>
</ROOT2>

我的可序列化代码如下：

[Serializable()]
[System.Xml.Serialization.XmlRoot("Root2")]
public class Root2
{
    [System.Xml.Serialization.XmlArray("Data2")]           
    [System.Xml.Serialization.XmlArrayItem("CUST_HOLD", typeof(CUST_HOLD))]
    public CUST_HOLD[] CUST_HOLD { get; set; }
    [System.Xml.Serialization.XmlArrayItem("HOLD", typeof(HOLD))]
    public HOLD[] HOLD { get; set; }    

}

我尝试了不同的配置，但这是唯一不会导致错误的配置。但问题是，只有第一个 XmlArrayItem 得到处理（在本例中为 CUST_HOLD）。另一个保持为空，而相应的数组中应该至少有一个项目。

4

2 回答 2

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如果你不能为它写一个 xsd，你就不能将它序列化/反序列化为 xml。

那将是 xsd 中任何序列的序列或极差的集合。

于 2012-09-24T16:10:39.237 回答
0

试试看，

代码

类声明

[Serializable()]
[System.Xml.Serialization.XmlInclude(typeof(CUST_HOLD))]
[System.Xml.Serialization.XmlInclude(typeof(HOLD))]
[System.Xml.Serialization.XmlType(TypeName = "Data2")]
public class Root2
{

    [System.Xml.Serialization.XmlArrayItem("CUST_HOLD")]
    public CUST_HOLD[] CUST_HOLD;


     [System.Xml.Serialization.XmlArrayItem("HOLD")]
    public HOLD[] HOLD;

}

 [Serializable()]
[System.Xml.Serialization.XmlType("CUST_HOLD")]
public class CUST_HOLD
{

    public int i;
}

[Serializable()]
[System.Xml.Serialization.XmlType("HOLD")]
public class HOLD
{

    public int i;
}

序列化

List<Root2> list = new List<Root2>();
Root2 obj = new Root2();
obj.CUST_HOLD = new CUST_HOLD[] { new CUST_HOLD() { i = 1 }, new CUST_HOLD() { i = 1 }, new CUST_HOLD() { i = 1 } };
obj.HOLD = new HOLD[] { new HOLD() { i = 1 }, new HOLD() { i = 1 }, new HOLD() { i = 1 } };

  list.Add(obj);

 //Serialize List<Root2>
 System.Xml.Serialization.XmlSerializer Serializer = new     System.Xml.Serialization.XmlSerializer(typeof(List<Root2>),new System.Xml.Serialization.XmlRootAttribute("Root2"));

 System.IO.MemoryStream mo = new System.IO.MemoryStream();
 Serializer.Serialize(mo, list);
 string str = UnicodeEncoding.UTF8.GetString(mo.ToArray());

输出：

<?xml version="1.0"?>
<Root2 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Data2>
<CUST_HOLD>
  <CUST_HOLD>
    <i>1</i>
  </CUST_HOLD>
  <CUST_HOLD>
    <i>1</i>
  </CUST_HOLD>
  <CUST_HOLD>
    <i>1</i>
  </CUST_HOLD>
</CUST_HOLD>
<HOLD>
  <HOLD>
    <i>1</i>
  </HOLD>
  <HOLD>
    <i>1</i>
  </HOLD>
  <HOLD>
    <i>1</i>
  </HOLD>
</HOLD>
</Data2>
</Root2>

反序列化

string str = @"<?xml version='1.0'?>
<Root2 xmlns:xsi='http://www.w3.org/2001/XMLSchema-instance'  xmlns:xsd='http://www.w3.org/2001/XMLSchema'>
<Data2>
<CUST_HOLD>
  <CUST_HOLD>
    <i>1</i>
  </CUST_HOLD>
  <CUST_HOLD>
    <i>1</i>
  </CUST_HOLD>
  <CUST_HOLD>
    <i>1</i>
  </CUST_HOLD>
</CUST_HOLD>
<HOLD>
  <HOLD>
    <i>1</i>
  </HOLD>
  <HOLD>
    <i>1</i>
  </HOLD>
  <HOLD>
    <i>1</i>
  </HOLD>
</HOLD>
</Data2>
</Root2>";

System.Xml.Serialization.XmlSerializer Serializer = new System.Xml.Serialization.XmlSerializer(typeof(List<Root2>), new System.Xml.Serialization.XmlRootAttribute("Root2"));

System.IO.MemoryStream mo = new   System.IO.MemoryStream(UnicodeEncoding.UTF8.GetBytes(str));

List<Root2> list = (List<Root2>)Serializer .Deserialize(mo);

注意 <Data2>标签仅在序列化类型List<Root2>为单Root2实例<Data2>标签时才会出现在 xml 字符串中，并且无法进行反序列化。反序列化将为CUST_HOLD&返回空值HOLD。返回类型应该是List<Root2>.

于 2012-09-24T17:47:25.570 回答