我正在尝试反序列化一些xml。这是:
<FooBars xmlns="http://foos">
<Id1 xmlns="http://bars">2</Id1>
<Id2 xmlns="http://bars">7</Id2>
<Info xmlns="http://bars">
<Data>
<Field1>text1</Field1>
<Field2>text2</Field2>
<Field3>text3</Field3>
</Data>
<Data>
<Field1>text5</Field1>
<Field2>text6</Field2>
<Field3>text7</Field3>
</Data>
</Info>
</FooBars>
我试过了:
var myData = (FooBars)serializer.Deserialize(foobars.CreateReader());
...
[XmlRoot(ElementName = "FooBars", Namespace = "http://foos")]
public class FooBars
{
[XmlElement(ElementName = "Id1", Namespace = "http://bars")]
public string Id1 { get; set; }
[XmlElement(ElementName = "Id2", Namespace = "http://bars")]
public string Id2 { get; set; }
[XmlElement(ElementName = "Info", Namespace = "http://bars")]
public List<Data> Info { get; set; }
}
public class Data
{
[XmlElement(ElementName = "Field1")]
public string Field1 { get; set; }
[XmlElement(ElementName = "Field2")]
public string Field2 { get; set; }
}
但看起来Data类不被视为xml的一部分,因为它无法读取它。我得到所有其他元素(ids)但不是数据中定义的东西。 我哪里错了?
答案 0 :(得分:1)
假定丢失</Info>
标记是一个错字,您只需要XmlArray
和XmlArrayItem
[XmlRoot(ElementName = "FooBars", Namespace = "http://foos")]
public class FooBars
{
[XmlElement(ElementName = "Id1", Namespace = "http://bars")]
public string Id1 { get; set; }
[XmlElement(ElementName = "Id2", Namespace = "http://bars")]
public string Id2 { get; set; }
[XmlArray(ElementName = "Info", Namespace = "http://bars"), XmlArrayItem("Data")] //<--
public List<Data> Info { get; set; }
}
答案 1 :(得分:0)
XML缺少Info
的结束标记。此外,您需要在Field3
类中定义Data
属性,并添加命名空间&#39; http://bars&#39;它。
答案 2 :(得分:0)
如果您的xml缺少信息结束标记,意图包含数据元素,那么您的类应该类似于:
[XmlRoot(ElementName = "FooBars", Namespace = "http://foos")]
public class FooBars
{
[XmlElement(ElementName = "Id1", Namespace = "http://bars")]
public string Id1 { get; set; }
[XmlElement(ElementName = "Id2", Namespace = "http://bars")]
public string Id2 { get; set; }
[XmlElement(ElementName = "Info", Namespace = "http://bars")]
public Info Information { get;set; }
}
public class Info {
[XmlElement(ElementName = "Data", Namespace = "")]
public Info[] Info { get; set; }
}
public class Data
{
[XmlElement(ElementName = "Field1")]
public string Field1 { get; set; }
[XmlElement(ElementName = "Field2")]
public string Field2 { get; set; }
}
请注意,在xml中,info对象基本上包含所有数据,而不是你设计类的FooBars。
答案 3 :(得分:0)
通过LinqPad - 您可以在myData.Info
属性中看到Field属性为null,这是您的问题,对吧? <强>已更新强>:
void Main()
{
string xmlString;
string path = @"C:\Temp\exampleXmlSO.xml";
using (StreamReader streamReader = File.OpenText(path))
{
xmlString = streamReader.ReadToEnd();
}
XmlSerializer serializer = new XmlSerializer(typeof(FooBars));
using (StringReader stringReader = new StringReader(xmlString))
{
var myData = (FooBars)serializer.Deserialize(stringReader);
Console.WriteLine(myData);
}
}
[XmlRoot(ElementName = "FooBars", Namespace = "http://foos")]
public class FooBars
{
[XmlElement(ElementName = "Id1", Namespace = "http://bars")]
public string Id1 { get; set; }
[XmlElement(ElementName = "Id2", Namespace = "http://bars")]
public string Id2 { get; set; }
[XmlArray(ElementName = "Info", Namespace = "http://bars"), XmlArrayItem("Data")] //<--
public List<Data> Info { get; set; }
}
[Serializable]
public class Data
{
[XmlElement(ElementName = "Field1")]
public string Field1 { get; set; }
[XmlElement(ElementName = "Field2")]
public string Field2 { get; set; }
[XmlIgnore]
public string Field3 { get; set; }
}