Question

我尝试将xml序列化到我的类：

[XmlRoot(ElementName = "order")]
public class OrderModel
{
    [XmlElement(ElementName = "purchaser")]
    public CompanyModel Purchaser { get; set; }

    [XmlElement(ElementName = "deliverver")]
    public CompanyModel Deliverver { get; set; }

    [XmlElement(ElementName = "position")]
    public List<OrderPositionModel> Positions { get; set; }
}

public class OrderPositionModel
{
    [XmlElement(ElementName = "id")]
    public int id { get; set; }

    [XmlElement(ElementName = "lp")]
    public int OrdinalNumber { get; set; }

    [XmlElement(ElementName = "name")]
    public string Name { get; set; }

    [XmlElement(ElementName = "quantity")]
    public int Quantity { get; set; }
}

序列化后，我得到具有良好List长度的对象，但其元素具有空id，lp，name和quantitiy。

更新序列化方法

private object XmlReader(HttpPostedFileBase file, Type type)
    {
        var document = new XmlDocument();
        document.Load(file.InputStream);

        string stream = document.OuterXml;

        StringReader strReader = null;
        XmlSerializer serializer = null;
        XmlTextReader xmlReader = null;
        Object obj = null;

        try
        {
            strReader = new StringReader(stream);
            serializer = new XmlSerializer(type);
            xmlReader = new XmlTextReader(strReader);
            obj = serializer.Deserialize(xmlReader);
        }
        catch (Exception)
        {
            //Handle Exception Code
        }
        finally
        {
            if (xmlReader != null)
            {
                xmlReader.Close();
            }
            if (strReader != null)
            {
                strReader.Close();
            }
        }
        return obj;
    }

像这样，然后我添加使用块

更新＃2 我试试这个xml

<order>
<purchaser>
    <name>name1</name>
    <address>address1</address>
</purchaser>
<deliverver>
    <name>name2</name>
    <address>address2</address>
</deliverver>
<position id="17" name="pos1" lp="1" quantity="1"/>
<position id="21" name="pos2" lp="2" quantity="2"/>
</order>

Answer 1

最简单的工作示例：

请注意，您的某些属性在使用XMLElement时属于其属性，因此应使用XMLAttribute进行修饰。

（我还删除了所有不必要的东西）

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

namespace ConsoleApplication1
{
    internal class Program
    {
        private static void Main(string[] args)
        {
            var filename = @"..\..\XMLFile1.xml";
            var model = Read(filename);
        }

        private static OrderModel Read(string filename)
        {
            using (var reader = XmlReader.Create(filename))
            {
                var serializer = new XmlSerializer(typeof(OrderModel));
                var model = (OrderModel) serializer.Deserialize(reader);
                return model;
            }
        }
    }

    [XmlRoot(ElementName = "order")]
    public class OrderModel
    {
        [XmlElement(ElementName = "purchaser")]
        public CompanyModel Purchaser { get; set; }

        [XmlElement(ElementName = "deliverver")]
        public CompanyModel Deliverver { get; set; }

        [XmlElement(ElementName = "position")]
        public List<OrderPositionModel> Positions { get; set; }
    }

    public class OrderPositionModel
    {
        [XmlAttribute("id")]
        public int id { get; set; }

        [XmlAttribute("lp")]
        public int OrdinalNumber { get; set; }

        [XmlAttribute("name")]
        public string Name { get; set; }

        [XmlAttribute("quantity")]
        public int Quantity { get; set; }
    }

    public class CompanyModel

    {
        [XmlElement(ElementName = "name")]
        public string Name { get; set; }

        [XmlElement(ElementName = "address")]
        public string Address { get; set; }
    }
}

Answer 2

尝试以这种方式进行序列化，它对我有用

       Example of using the class "ObjectSerialization"

        try
        {
            parmExtRoute viagens = new parmExtRoute();
            Stream streamXml = arquivoRecebido.InputStream;
            var document = new XmlDocument();
            document.Load(streamXml);

            string stream = document.OuterXml;
            viagens = ObjectSerialization.DeserializeObject<parmExtRoute>(stream);
            retorno.Result = viagens;
        }
        catch (Exception ex)
        {
            retorno.ListErrors.Add(ex.Message);
        }

        return retorno;
    



public static class ObjectSerialization
{
    public static string SerializeObject<T>(T dataObject)
    {
        if (dataObject == null)
            return string.Empty;

        try
        {
            using (var stringWriter = new StringWriter())
            {
                var namespaces = new XmlSerializerNamespaces();
                var settings = new XmlWriterSettings();
                settings.Encoding = new UnicodeEncoding(false, false);
                settings.Indent = false;
                settings.NewLineChars = string.Empty;
                settings.NewLineHandling = NewLineHandling.None;
                settings.OmitXmlDeclaration = true;
                namespaces.Add(string.Empty, string.Empty);
                var serializer = new XmlSerializer(typeof(T));

                using (var xmlWriter = XmlWriter.Create(stringWriter, settings))
                    serializer.Serialize(xmlWriter, dataObject, namespaces);

                return stringWriter.ToString();
            }
        }
        catch (Exception ex)
        {
            Console.WriteLine(string.Format("Erro ao serializar objeto. ", ex.Message) + DateTime.Now.ReturnDateLocale());
            return string.Empty;
        }
    }
    public static T DeserializeObject<T>(string xml) where T : class
    {
        if (string.IsNullOrEmpty(xml))
            return Activator.CreateInstance<T>();

        try
        {
            using (var stringReader = new StringReader(xml))
            {
                var serializer = new XmlSerializer(typeof(T), new Type[] { typeof(T) }); ;
                return (T)serializer.Deserialize(stringReader);
            }
        }
        catch (Exception ex)
        {
            Console.WriteLine(string.Format("Erro ao deserializar objeto. ", ex.Message) + DateTime.Now.ReturnDateLocale());
            return Activator.CreateInstance<T>();
        }
    }
    public static string ObjectParamWs<T>(T dataObject)
    {
        string conteudo = SerializeObject<T>(dataObject);
        string xml = conteudo;
        return xml;
    }
}

xml序列化后具有空属性的项列表

2 个答案: