我正在尝试反序列化以下XML:
<?xml version="1.0" encoding="UTF-8"?>
<XGResponse><Failure code="400">
Message id '1' was already submitted.
</Failure></XGResponse>通过此调用:
[...]
var x = SerializationHelper.Deserialize<XMLGateResponse.XGResponse>(nResp);
[...]
public static T Deserialize<T>(string xml)
{
using (var str = new StringReader(xml))
{
var xmlSerializer = new XmlSerializer(typeof(T));
return (T)xmlSerializer.Deserialize(str);
}
}要获取相应类的实例,请执行以下操作:
//------------------------------------------------------------------------------
// <auto-generated>
// This code was generated by a tool.
// Runtime Version:4.0.30319.18052
//
// Changes to this file may cause incorrect behavior and will be lost if
// the code is regenerated.
// </auto-generated>
//------------------------------------------------------------------------------
using System.Xml.Serialization;
//
// This source code was auto-generated by xsd, Version=4.0.30319.1.
//
namespace XMLGateResponse
{
/// <remarks/>
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.1")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "http://tempuri.org/XMLGateResponse")]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://tempuri.org/XMLGateResponse", IsNullable = false)]
public partial class XGResponse
{
private object[] itemsField;
/// <remarks/>
[System.Xml.Serialization.XmlElementAttribute("Failure", typeof(Failure))]
[System.Xml.Serialization.XmlElementAttribute("Success", typeof(Success))]
public object[] Items
{
get
{
return this.itemsField;
}
set
{
this.itemsField = value;
}
}
}
/// <remarks/>
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.1")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "http://tempuri.org/XMLGateResponse")]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://tempuri.org/XMLGateResponse", IsNullable = false)]
public partial class Failure
{
private string codeField;
private string titleField;
private string valueField;
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute(DataType = "NMTOKEN")]
public string code
{
get
{
return this.codeField;
}
set
{
this.codeField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string title
{
get
{
return this.titleField;
}
set
{
this.titleField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlTextAttribute()]
public string Value
{
get
{
return this.valueField;
}
set
{
this.valueField = value;
}
}
}
/// <remarks/>
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.1")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "http://tempuri.org/XMLGateResponse")]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://tempuri.org/XMLGateResponse", IsNullable = false)]
public partial class Success
{
private string titleField;
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string title
{
get
{
return this.titleField;
}
set
{
this.titleField = value;
}
}
}
}但它会引发错误There is an error in XML document (2, 2)。
我已经为这个问题寻找了大约一个小时的解决方案,但没有太大帮助。
我甚至尝试了一个不应该做任何事情的轻微更改:
public static T Deserialize<T>(string xml)
{
[...]
var xmlSerializer = new XmlSerializer(typeof(T), new XmlRootAttribute(typeof(T).Name));
[...]
}然而,这确实防止了错误的发生。但由于它只实现了返回一个完全为空的XMLGateResponse.XGResponse实例(每个元素/属性都为空),所以这并不是真正的改进。
我知道这种问题There is an error in XML document (2, 2)已经讨论了很多,但我真的没有找到一个适合我的解决方案。
发布于 2013-08-22 18:20:05
用指定默认名称空间名称的XmlRootAttribute修饰XGResponse,但您的文档没有这样做。
删除此命名空间声明或将xmlns="http://tempuri.org/XMLGateResponse"添加到xml的根元素中。
发布于 2014-03-06 15:58:42
如果您试图反序列化到错误的类型,您可能会得到相同的错误。
例如,如果您调用
Deserialize<object>(myXml)或
Deserialize<Failure>(myXml)我知道在以下情况下回答问题是一种糟糕的做法:1)答案已经提供,2)答案并不完全是提问者所要求的;但我认为这可能会节省一些时间,让其他人在这里找到与提问者不完全相同的问题。
https://stackoverflow.com/questions/18377554
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