Question

我有下一个xml：

<table-display-fields>
  <field name="NAME/>
  <field name="DESCRIPTION" />
</table-display-fields>

我用下一个代码实现了这个目标：

[XmlArray("table-display-fields")]
[XmlArrayItem("field")]
public TableDisplayField[] TableDisplayFields;

然后我将新的xml元素添加到table-display-fields节点：

<table-display-fields>
  <record-number-field name="ID" />
  <field name="NAME/>
  <field name="DESCRIPTION" />
</table-display-fields>

然后添加下一个代码来deserealize record-number-field：

[XmlArray("table-display-fields")]
[XmlArrayItem("record-number-field")]
public TableDisplayField[] RecordTableDisplayFields;

[XmlArray("table-display-fields")]
[XmlArrayItem("field")]
public TableDisplayField[] TableDisplayFields;

这不起作用。如何取消实现新的xml，并保存现有的属性路径？

Answer 1

您必须删除XmlArrayItem（）属性。

[XmlArray("table-display-fields")] 
public object[] TableDisplayItems;

TableDisplayItems中的每个对象都是field或record-number-field。

当然，如果您的阵列上只有一个record-number-field，那么解决方案可以更好。是这样的吗？

Answer 2

如果您能够对您的要求进行更详细的描述，那么将会有所帮助。如果您正在寻找正常的序列化/反序列化，您可以尝试以下解决方案：

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

namespace Serialization01 {
    [XmlRoot( "table-display-fields" )]
    public class TableDisplayFields {
        [XmlElement( "record-number-field" )]
        public string RecordNumberField { get; set; }

        [XmlElement( "field" )]
        public List<string> FieldName { get; set; }

        public TableDisplayFields ( ) {
            FieldName = new List<string>( 5 );
        }
    }
}

并使用以下代码编写和读取序列化数据：

using System.IO;
using System.Xml.Serialization;
using System;

namespace Serialization01 {
    class Program {
        static void Main ( string [] args ) {
            // Initiate the class
            TableDisplayFields t = new TableDisplayFields( );

            t.RecordNumberField = "ID";
            t.FieldName.Add( "NAME" );
            t.FieldName.Add( "DESCRIPTION" );

            TextWriter tw = new StreamWriter( Path.Combine( Environment.CurrentDirectory, "Data.xml" ) );

            XmlSerializer xs = new XmlSerializer( t.GetType( ) );

            xs.Serialize( tw, t );

            tw.Flush( );
            tw.Close( );

            TextReader tr = new StreamReader( Path.Combine( Environment.CurrentDirectory, "Data.xml" ) );
            TableDisplayFields t2 = xs.Deserialize( tr ) as TableDisplayFields;

            Console.WriteLine( "RecordNumberField for t2 is {0}", t2.RecordNumberField );
            foreach ( string field in t2.FieldName ) {
                Console.WriteLine( "Found field '{0}'", field );
            }
        }
    }
}

希望有所帮助。

将xml不同的元素序列化为c＃中的几个属性

2 个答案: