Question

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

XML基本上是这样的：

<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，但这是唯一一个不会导致错误的配置。但问题是，只有第一个XmlArrayItem被处理（在本例中为CUST_HOLD）。另一个保持为null，而相应数组中至少应有一个项目。

Answer 1

如果您无法为其编写xsd，则无法将其序列化/反序列化为xml。

这将是xsd中的任何序列或极其糟糕的集合。

Answer 2

试试，

<强> CODE

课程声明

[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;
}

<强>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());

<强>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>

<强>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>标记仅在单个List<Root2>实例Root2标记的序列化类型为<Data2>时才会显示在xml字符串中并且不会反序列化做不到。反序列化将返回CUST_HOLD＆amp;的空值。 HOLD。返回类型应为List<Root2>。

c＃xml可序列化的多个arrayitems

2 个答案: