Question

我正在尝试阅读一个简单的xml并通过类反序列化它。但它返回xml中存在的所有参数的空值。在反序列化以下xml时，任何人都可以帮我解决空的返回值问题。

<Batch>
    <description>example</description>
    <fileField>DesktopA</fileField>
    <output>The</output>
    <input>home</input>
    <input>green</input>
    <parameters>
        <action>1</action>
        <item>2</item>          
    </parameters>
    <parameters>
        <action>1</action>
        <item>4</item>          
    </parametersField>
</Batch>

班级：

[Serializable()]
public partial class Batch {

    private string description;

    private string fileField;

    private string output;

    private string[] input;

    private Parameter[] parameters;

    /// <remarks/>
    public string Description {
        get {
            return this.description;
        }
        set {
            this.description = value;
        }
    }

    /// <remarks/>
    public string File {
        get {
            return this.fileField;
        }
        set {
            this.fileField = value;
        }
    }

    /// <remarks/>
    public string Output {
        get {
            return this.output;
        }
        set {
            this.output = value;
        }
    }

    /// <remarks/>
    [XmlElement]
    public string[] Input {
        get {
            return this.input;
        }
        set {
            this.input = value;
        }
    }

   /// <remarks/>
   [XmlElement]
   public Parameter[] Parameters {
        get {
            return this.parameters;
        }
        set {
            this.parameters = value;
        }
    }
}

/// <remarks/>

public partial class Parameter {


    private sbyte action;  

    private object item;



    /// <remarks/>
    public sbyte Action {
        get {
            return this.action;
        }
        set {
            this.action = value;
        }
    }

   /// <remarks/>
   [XmlElement]
   public object Item {
        get {
            return this.item;
        }
        set {
            this.item = value;
        }
    }

deserializaion方法：

XmlSerializer reader;
                Type bt = batchInfo.GetType();

                reader = new XmlSerializer(bt);                 
                System.IO.StreamReader file =
                   new System.IO.StreamReader(testScriptFile, System.Text.Encoding.Default);                
            batchInfo = (Batch)reader.Deserialize(file);

Answer 1

如果您正确检查了XML，则可以看到参数存在问题。你有，但它应该是

<Batch>
    <description>example</description>
    <fileField>DesktopA</fileField>
    <output>The</output>
    <input>home</input>
    <input>green</input>
    <parameters>
        <action>1</action>
        <item>2</item>          
    </parameters>
    <parameters>
        <action>1</action>
        <item>4</item>          
    </parameters>
</Batch>

希望它有所帮助！

Answer 2

您可以使用以下模型反序列化您的xml。它使用XmlElement属性。

public class Parameter
{
    [XmlElement("action")]
    public int Action { get; set; }
    [XmlElement("item")]
    public int Item { get; set; }
}

[XmlRoot("Batch")]
public class Batch
{
    [XmlElement("description")]
    public string Description { get; set; }
    [XmlElement("fileField")]
    public string FileField { get; set; }
    [XmlElement("output")]
    public string Output { get; set; }
    [XmlElement("input")]
    public List<string> Inputs { get; set; }
    [XmlElement("parameters")]
    public List<Parameter> Parameters { get; set; }
}

通过使用此代码，您可以执行反序列化。

var xml = @"
    <Batch>
        <description>example</description>
        <fileField>DesktopA</fileField>
        <output>The</output>
        <input>home</input>
        <input>green</input>
        <parameters>
            <action>1</action>
            <item>2</item>          
        </parameters>
        <parameters>
            <action>1</action>
            <item>4</item>          
        </parameters>
    </Batch>";

var serializer = new XmlSerializer(typeof(Batch));
var textReader = new StringReader(xml);
var obj = (Batch) serializer.Deserialize(textReader);

或者从文件中读取

var serializer = new XmlSerializer(typeof(Batch));
using (var fileStream = new FileStream(@"C:\mypath\myFile.xml", FileMode.Open))
{
    var obj = (Batch)serializer.Deserialize(fileStream);
}

反序列化简单的xml返回空值

2 个答案: