【发布时间】:2019-11-16 20:29:58
【问题描述】:
我从这里复制了 xml 文档How to Deserialize XML document
<?xml version="1.0" encoding="utf-8"?>
<CarCollection>
<Cars>
<Car>
<StockNumber>1020</StockNumber>
<Make>Nissan</Make>
<Model>Sentra</Model>
</Car>
<Car>
<StockNumber>1010</StockNumber>
<Make>Toyota</Make>
<Model>Corolla</Model>
</Car>
<Car>
<StockNumber>1111</StockNumber>
<Make>Honda</Make>
<Model>Accord</Model>
</Car>
</Cars>
</CarCollection>
我将该代码与从 /paste special/paste xml 生成的类一起用作类
using System;
using System.IO;
using System.Xml.Serialization;
namespace DeSerialize
{
class Program
{
static void Main(string[] args)
{
XmlSerializer serializer =
new XmlSerializer(typeof(CarCollectionCar));
// Declare an object variable of the type to be deserialized.
CarCollectionCar i;
using (Stream reader = new FileStream("cars.xml", FileMode.Open))
{
// Call the Deserialize method to restore the object's state.
i = (CarCollectionCar)serializer.Deserialize(reader);
}
// Write out the properties of the object.
Console.Write(
// i.StockNumber + "\t" +
i.StockNumber + "\t" +
//i.StockNumber + "\t" +
i.Model + "\t" +
i.Make);
}
}
[Serializable()]
public partial class CarCollection
{
/// <remarks/>
[XmlArrayItem("Car", IsNullable = false)]
public CarCollectionCar[] Cars { get; set; }
}
/// <remarks/>
[Serializable()]
[System.ComponentModel.DesignerCategory("code")]
[XmlType(AnonymousType = true)]
public partial class CarCollectionCar
{
/// <remarks/>
public ushort StockNumber { get; set; }
/// <remarks/>
public string Make { get; set; }
/// <remarks/>
public string Model { get; set; }
}
}
我得到错误
Unhandled Exception: System.InvalidOperationException: There is an error in XML
document (2, 2). ---> System.InvalidOperationException: <CarCollection xmlns=''>
was not expected.
at
Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReaderCarCollectionCar.Read3_CarCollectionCar()
--- End of inner exception stack trace ---
at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle, XmlDeserializationEvents events)
at System.Xml.Serialization.XmlSerializer.Deserialize(Stream stream)
at DeSerialize.Program.Main(String[] args)
如何解决问题并输出所需的 Cars 参数?
【问题讨论】:
-
这是一个专业提示。尝试先序列化您的课程。然后看看反序列化有什么问题。
-
new XmlSerializer(typeof(CarCollectionCar));应该是new XmlSerializer(typeof(CarCollection));,因为您正在反序列化汽车集合,而不是单个汽车。
标签: c# xml serialization