Question

让我将XML转换为C＃中的对象变得更加复杂。

我想转换一个由对象列表组成的XML＆＃39; Regla＆＃39;包含一系列字段（idRegla，DateFrom，DateTo和可能未显示的异常列表）。

我疯了，我认为这并不难......

这是XML：

<ListaReglas>
  <REGLA>
    <idRegla>2</idRegla>
    <DateFrom>2013-12-01T00:00:00</DateFrom>
    <DateTo>2015-07-25T00:00:00</DateTo>
    <Excepciones>
      <FECHA>2013-12-25T00:00:00</FECHA>
    </Excepciones>
  </REGLA>
  <REGLA>
    <idRegla>4</idRegla>
    <DateFrom>2013-12-01T00:00:00</DateFrom>
    <DateTo>2015-07-25T00:00:00</DateTo>
    <Excepciones>
      <FECHA>2013-12-25T00:00:00</FECHA>
    </Excepciones>
  </REGLA>
  <REGLA>
    <idRegla>5</idRegla>
    <DateFrom>2013-12-01T00:00:00</DateFrom>
    <DateTo>2015-07-25T00:00:00</DateTo>
    <Excepciones>
      <FECHA>2013-12-25T00:00:00</FECHA>
    </Excepciones>
  </REGLA>
  <REGLA>
    <idRegla>7</idRegla>
    <DateFrom>2013-11-19T00:00:00</DateFrom>
    <DateTo>2015-12-19T00:00:00</DateTo>
  </REGLA>
</ListaReglas>

这是我的班级：

        [Serializable]
        [XmlTypeAttribute(AnonymousType = true)]
        public class ReglaRangoResult
        {
            [XmlElement(ElementName = "idRegla", IsNullable = false)]
            public int idRegla { get; set; }

            [XmlElement(ElementName = "DateFrom", IsNullable = false)]
            public DateTime DateFrom { get; set; }

            [XmlElement(ElementName = "DateTo", IsNullable = false)]
            public DateTime DateTo { get; set; }

            [XmlElement(ElementName = "Excepciones", IsNullable = true)]
            public List<DateTime> Excepciones { get; set; }

            [XmlIgnore]
            public int Peso { get; set; }
        }

这是我的代码：

       [...]
       List<ReglaRangoResult> listaReglas = new List<ReglaRangoResult>();
       XmlDoc xmlDoc = new XmlDoc(rdr.GetString(0));

       foreach (XmlNode xmlNode in xmlDoc.SelectNodes("//ListaReglas/REGLA"))
       {
            listaReglas.Add(XmlToObject<ReglaRangoResult>(xmlNode.OuterXml));
       }
       [...]


        public static T XmlToObject<T>(string xml)
        {
            using (var xmlStream = new StringReader(xml))
            {
                var serializer = new XmlSerializer(typeof(T));
                return (T)serializer.Deserialize(XmlReader.Create(xmlStream));
            }
        }

我不明白我做错了什么。 ReglaRangoResult是否配置错误？缺什么？还剩下什么？

返回异常：

＆＃39;错误反映类型＆＃39; dllReglasNegocioMP.ReglaRangoResult＆＃39;

Answer 1

在Visual Studio 2013中，您可以使用XML并选择“编辑/粘贴特殊/粘贴XML作为类”。完成后，您可以使用XmlSerializer以简单的方式序列化和反序列化。

 var serializer = new System.Xml.Serialization.XmlSerializer(typeof(MyPastedClass));
 MyPastedClass obj;
 using (var xmlStream = new StringReader(str))
 {
      obj = (MyPastedClass)serializer.Deserialize(xmlStream);
 }

Answer 2

参加下面列出的课程。您可以将对象序列化为真实XML并进行比较。

using System.Diagnostics.Contracts;
using System.Globalization;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;

namespace VmsSendUtil
{
    /// <summary> Serializes and Deserializes any object to and from string </summary>
    public static class StringSerializer
    {
        ///<summary> Serializes object to string </summary>
        ///<param name="obj"> Object to serialize </param>
        ///<returns> Xml string with serialized object </returns>
        public static string Serialize<T>(T obj)
        {
            Contract.Ensures(!string.IsNullOrEmpty(Contract.Result<string>()));

            var stringSerializer = new StringSerializer<T>();
            return stringSerializer.Serialize(obj);
        }

        /// <summary> Deserializes object from string. </summary>
        /// <param name="xml"> String with serialization XML data </param>
        public static T Deserialize<T>(string xml)
        {
            Contract.Requires(!string.IsNullOrEmpty(xml));
            Contract.Ensures(!Equals(Contract.Result<T>(), null));

            var stringSerializer = new StringSerializer<T>();
            return stringSerializer.Deserialize(xml);
        }
    }

    /// <summary> Serializes and Deserializes any object to and from string </summary>
    public class StringSerializer<T>
    {
        [ContractInvariantMethod]
        private void ObjectInvariant()
        {
            Contract.Invariant(_serializer != null);
        }

        private readonly XmlSerializer _serializer = new XmlSerializer(typeof(T));

        ///<summary> Serializes object to string </summary>
        ///<param name="obj"> Object to serialize </param>
        ///<returns> Xml string with serialized object </returns>
        public string Serialize(T obj)
        {
            Contract.Ensures(!string.IsNullOrEmpty(Contract.Result<string>()));

            var sb = new StringBuilder();
            using (var sw = new StringWriter(sb, CultureInfo.InvariantCulture))
            {
                var tw = new XmlTextWriter(sw) { Formatting = Formatting.Indented };
                _serializer.Serialize(tw, obj);
            }
            string result = sb.ToString();
            Contract.Assume(!string.IsNullOrEmpty(result));
            return result;
        }

        /// <summary> Deserializes object from string. </summary>
        /// <param name="xml"> String with serialization XML data </param>
        public T Deserialize(string xml)
        {
            Contract.Requires(!string.IsNullOrEmpty(xml));
            Contract.Ensures(!Equals(Contract.Result<T>(), null));

            using (var stringReader = new StringReader(xml))
            {
                // Switch off CheckCharacters to deserialize special characters
                var xmlReaderSettings = new XmlReaderSettings { CheckCharacters = false };

                var xmlReader = XmlReader.Create(stringReader, xmlReaderSettings);
                var result = (T)_serializer.Deserialize(xmlReader);
                Contract.Assume(!Equals(result, null));
                return result;
            }
        }
    }
}

Answer 3

您将获得一个异常，该异常具有您未在代码中定义的层次结构。如果您设置了正确的层次结构，它将起作用。

将XML转换为c＃对象

3 个答案: