我如何反序列化这个XML文档:

<?xml version="1.0" encoding="utf-8"?>
<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>

我有这个:

[Serializable()]
public class Car
{
    [System.Xml.Serialization.XmlElementAttribute("StockNumber")]
    public string StockNumber{ get; set; }

    [System.Xml.Serialization.XmlElementAttribute("Make")]
    public string Make{ get; set; }

    [System.Xml.Serialization.XmlElementAttribute("Model")]
    public string Model{ get; set; }
}

.

[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
    [XmlArrayItem(typeof(Car))]
    public Car[] Car { get; set; }

}

.

public class CarSerializer
{
    public Cars Deserialize()
    {
        Cars[] cars = null;
        string path = HttpContext.Current.ApplicationInstance.Server.MapPath("~/App_Data/") + "cars.xml";

        XmlSerializer serializer = new XmlSerializer(typeof(Cars[]));

        StreamReader reader = new StreamReader(path);
        reader.ReadToEnd();
        cars = (Cars[])serializer.Deserialize(reader);
        reader.Close();

        return cars;
    }
}

这似乎并不奏效:-(


当前回答

使用泛型类来反序列化XML文档如何

//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
// Generic class to load any xml into a class
// used like this ...
// YourClassTypeHere InfoList = LoadXMLFileIntoClass<YourClassTypeHere>(xmlFile);

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

public static T LoadXMLFileIntoClass<T>(string xmlFile)
{
    T returnThis;
    XmlSerializer serializer = new XmlSerializer(typeof(T));
    if (!FileAndIO.FileExists(xmlFile))
    {
        Console.WriteLine("FileDoesNotExistError {0}", xmlFile);
    }
    returnThis = (T)serializer.Deserialize(new StreamReader(xmlFile));
    return (T)returnThis;
}

这一部分可能是必要的,也可能不是。在Visual Studio中打开XML文档,右键单击XML,选择属性。然后选择您的模式文件。

其他回答

下面的代码段应该可以做到这一点(你可以忽略大部分序列化属性):

public class Car
{
  public string StockNumber { get; set; }
  public string Make { get; set; }
  public string Model { get; set; }
}

[XmlRootAttribute("Cars")]
public class CarCollection
{
  [XmlElement("Car")]
  public Car[] Cars { get; set; }
}

...

using (TextReader reader = new StreamReader(path))
{
  XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
  return (CarCollection) serializer.Deserialize(reader);
}

使用泛型类来反序列化XML文档如何

//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
// Generic class to load any xml into a class
// used like this ...
// YourClassTypeHere InfoList = LoadXMLFileIntoClass<YourClassTypeHere>(xmlFile);

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

public static T LoadXMLFileIntoClass<T>(string xmlFile)
{
    T returnThis;
    XmlSerializer serializer = new XmlSerializer(typeof(T));
    if (!FileAndIO.FileExists(xmlFile))
    {
        Console.WriteLine("FileDoesNotExistError {0}", xmlFile);
    }
    returnThis = (T)serializer.Deserialize(new StreamReader(xmlFile));
    return (T)returnThis;
}

这一部分可能是必要的,也可能不是。在Visual Studio中打开XML文档,右键单击XML,选择属性。然后选择您的模式文件。

看看这是否有帮助:

[Serializable()]
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
    [XmlArrayItem(typeof(Car))]
    public Car[] Car { get; set; }
}

.

[Serializable()]
public class Car
{
    [System.Xml.Serialization.XmlElement()]
    public string StockNumber{ get; set; }

    [System.Xml.Serialization.XmlElement()]
    public string Make{ get; set; }

    [System.Xml.Serialization.XmlElement()]
    public string Model{ get; set; }
}

如果做不到这一点,请使用visual studio附带的xsd.exe程序来基于该xml文件创建一个模式文档,然后再次使用它来基于该模式文档创建一个类。

如果您的.xml文件已经在磁盘的某个地方生成,并且您使用了List<T>,请尝试此代码块:

//deserialization

XmlSerializer xmlser = new XmlSerializer(typeof(List<Item>));
StreamReader srdr = new StreamReader(@"C:\serialize.xml");
List<Item> p = (List<Item>)xmlser.Deserialize(srdr);
srdr.Close();`

注意:C:\serialize.xml是我的.xml文件的路径。你可以根据自己的需要进行更改。

这是一个工作版本。我将XmlElementAttribute标签更改为XmlElement,因为在xml中,StockNumber、Make和Model值是元素,而不是属性。我还删除了reader.ReadToEnd();(该函数读取整个流并返回一个字符串,因此Deserialize()函数不能再使用阅读器了…位置在小溪的尽头)。我在命名上也做了一些改动:)。

下面是这些类:

[Serializable()]
public class Car
{
    [System.Xml.Serialization.XmlElement("StockNumber")]
    public string StockNumber { get; set; }

    [System.Xml.Serialization.XmlElement("Make")]
    public string Make { get; set; }

    [System.Xml.Serialization.XmlElement("Model")]
    public string Model { get; set; }
}


[Serializable()]
[System.Xml.Serialization.XmlRoot("CarCollection")]
public class CarCollection
{
    [XmlArray("Cars")]
    [XmlArrayItem("Car", typeof(Car))]
    public Car[] Car { get; set; }
}

反序列化函数:

CarCollection cars = null;
string path = "cars.xml";

XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));

StreamReader reader = new StreamReader(path);
cars = (CarCollection)serializer.Deserialize(reader);
reader.Close();

和稍微调整的xml(我需要添加一个新元素来包装<Cars>…Net对反序列化数组很挑剔):

<?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>