Question

我正在尝试从Web服务反序列化响应。

性反应：

<?xml version="1.0" encoding="UTF-8"?>
  <prestashop xmlns:xlink="http://www.w3.org/1999/xlink">
    <currency>
      <id><![CDATA[1]]></id>
      <name><![CDATA[Złoty]]></name>
      <iso_code><![CDATA[PLN]]></iso_code>
      <iso_code_num><![CDATA[985]]></iso_code_num>
      <blank><![CDATA[1]]></blank>
      <sign><![CDATA[zł]]></sign>
      <format><![CDATA[2]]></format>
      <decimals><![CDATA[1]]></decimals>
      <conversion_rate><![CDATA[1.000000]]></conversion_rate>
      <deleted><![CDATA[0]]></deleted>
      <active><![CDATA[1]]></active>
    </currency>
  </prestashop>

我的班级代码：

public class prestashop
{
   Currency currency;
}

public class Currency
{
    public int id;
    public string name;
    public string iso_code;
    public int iso_code_num;
    public int blank;
    public string sign;
    public int format;
    public int decimals;
    public float conversion_rate;
    public bool deleted;
    public bool active;
}

最后我尝试反序列化的部分：

Prestashop_WebClient client = new Prestashop_WebClient();
prestashop objpresta = new prestashop();
StreamReader strumien = client.ReadRecord("currencies", "1");
XmlSerializer ser = new XmlSerializer(typeof(prestashop));
objpresta = (prestashop)ser.Deserialize(strumien);
strumien.Close();

当我运行代码时，反序列化后对象objpresta为null但没有错误。

我会非常感谢任何帮助。

Answer 1

试试这个......

Usings .....

using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;

类.....

[XmlRoot(ElementName = "currency")]
public class Currency
{
    [XmlElement(ElementName = "id")]
    public string Id { get; set; }
    [XmlElement(ElementName = "name")]
    public string Name { get; set; }
    [XmlElement(ElementName = "iso_code")]
    public string Iso_code { get; set; }
    [XmlElement(ElementName = "iso_code_num")]
    public string Iso_code_num { get; set; }
    [XmlElement(ElementName = "blank")]
    public string Blank { get; set; }
    [XmlElement(ElementName = "sign")]
    public string Sign { get; set; }
    [XmlElement(ElementName = "format")]
    public string Format { get; set; }
    [XmlElement(ElementName = "decimals")]
    public string Decimals { get; set; }
    [XmlElement(ElementName = "conversion_rate")]
    public string Conversion_rate { get; set; }
    [XmlElement(ElementName = "deleted")]
    public string Deleted { get; set; }
    [XmlElement(ElementName = "active")]
    public string Active { get; set; }
}

[XmlRoot(ElementName = "prestashop")]
public class Prestashop
{
    [XmlElement(ElementName = "currency")]
    public Currency Currency { get; set; }
    [XmlAttribute(AttributeName = "xlink", Namespace = "http://www.w3.org/2000/xmlns/")]
    public string Xlink { get; set; }
}

代码.....

        string strXML = File.ReadAllText("xml.xml");
        byte[] bufXML = ASCIIEncoding.UTF8.GetBytes(strXML);
        MemoryStream ms1 = new MemoryStream(bufXML);

        // Deserialize to object
        XmlSerializer serializer = new XmlSerializer(typeof(Prestashop));
        try
        {
            using (XmlReader reader = new XmlTextReader(ms1))
            {
                Prestashop deserializedXML = (Prestashop)serializer.Deserialize(reader);

            }// put a break point here and mouse-over Label1Text and Label2Text ….
        }
        catch (Exception ex)
        {
            throw;
        }

您的XML

<?xml version="1.0" encoding="UTF-8"?>
  <prestashop xmlns:xlink="http://www.w3.org/1999/xlink">
    <currency>
      <id><![CDATA[1]]></id>
      <name><![CDATA[Złoty]]></name>
      <iso_code><![CDATA[PLN]]></iso_code>
      <iso_code_num><![CDATA[985]]></iso_code_num>
      <blank><![CDATA[1]]></blank>
      <sign><![CDATA[zł]]></sign>
      <format><![CDATA[2]]></format>
      <decimals><![CDATA[1]]></decimals>
      <conversion_rate><![CDATA[1.000000]]></conversion_rate>
      <deleted><![CDATA[0]]></deleted>
      <active><![CDATA[1]]></active>
    </currency>
  </prestashop>

我正在从名为xml.xml的应用程序构建文件夹中的文件中读取XML到一个字符串...您需要从其他地方获取XML字符串或创建xml.xml文件并保存XML上面的代码工作

Answer 2

您正在拼写所有属性，就像XML元素一样，因此不需要为每个属性添加[XmlElement]。但我认为XmlRootAttribute缺失了：

[XmlRoot("currency")]
public class Currency
...

XmlSerializer.Deserialize（）保持返回null并且不会抛出任何错误

2 个答案: