如何反序列化XML文档

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:-(

780405 次浏览
小开

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。

小开
最佳答案

这是一个工作版本。我把XmlElementAttribute标签改为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(); (函数读取整个流并返回一个字符串，因此Deserialize()函数不能再使用读取器了……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wrap < car >…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>
小开

如果您只是将xml保存到一个文件中，然后使用xsd生成c#类呢?

  1. 将文件写入磁盘(我将其命名为foo.xml)
  2. 生成xsd: xsd foo.xml
  3. 生成c#: xsd foo.xsd /classes

等等，瞧——和c#代码文件，应该能够通过XmlSerializer读取数据:

    XmlSerializer ser = new XmlSerializer(typeof(Cars));
Cars cars;
using (XmlReader reader = XmlReader.Create(path))
{
cars = (Cars) ser.Deserialize(reader);
}

(将生成的foo.cs包含在项目中)

小开

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);
}
小开

我不认为.net对反序列化数组很挑剔。第一个xml文档格式不正确。 没有根元素，尽管看起来有。规范xml文档有一个根和至少一个元素(如果有的话)。在你的例子中:

<Root> <-- well, the root
<Cars> <-- an element (not a root), it being an array
<Car> <-- an element, it being an array item
...
</Car>
</Cars>
</Root>
小开

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。

小开

有两种可能。

方法1。# EYZ0工具

人力资源/ > < p > < 假设XML文件位于这个位置C:\path\to\xml\file.xml

    <李>打开# EYZ0 < br > 您可以在Start Menu > Programs > Microsoft Visual Studio 2012 > Visual Studio Tools中找到它 或者如果你有Windows 8，可以在开始屏幕中输入开发人员命令提示符
  1. 通过键入cd /D "C:\path\to\xml"将位置更改为XML文件目录
  2. 从xml文件中输入xsd file.xml创建XSD文件
  3. 输入xsd /c file.xsd创建c#类

就是这样!您已经从C:\path\to\xml\file.cs中的xml文件生成了c#类

方法2 -粘贴特殊

人力资源/ > < p > < 需要Visual Studio 2012+

  1. 将XML文件的内容复制到剪贴板
  2. 向您的解决方案添加新的空类文件(转变+Alt+C)
    3. 打开该文件，在菜单中单击Edit > Paste special > Paste XML As Classes
    enter image description here

就是这样!

使用

这个helper类的用法非常简单:

using System;
using System.IO;
using System.Web.Script.Serialization; // Add reference: System.Web.Extensions
using System.Xml;
using System.Xml.Serialization;


namespace Helpers
{
internal static class ParseHelpers
{
private static JavaScriptSerializer json;
private static JavaScriptSerializer JSON { get { return json ?? (json = new JavaScriptSerializer()); } }


public static Stream ToStream(this string @this)
{
var stream = new MemoryStream();
var writer = new StreamWriter(stream);
writer.Write(@this);
writer.Flush();
stream.Position = 0;
return stream;
}




public static T ParseXML<T>(this string @this) where T : class
{
var reader = XmlReader.Create(@this.Trim().ToStream(), new XmlReaderSettings() { ConformanceLevel = ConformanceLevel.Document });
return new XmlSerializer(typeof(T)).Deserialize(reader) as T;
}


public static T ParseJSON<T>(this string @this) where T : class
{
return JSON.Deserialize<T>(@this.Trim());
}
}
}

你现在要做的就是:

    public class JSONRoot
{
public catalog catalog { get; set; }
}
// ...


string xml = File.ReadAllText(@"D:\file.xml");
var catalog1 = xml.ParseXML<catalog>();


string json = File.ReadAllText(@"D:\file.json");
var catalog2 = json.ParseJSON<JSONRoot>();
小开

如果使用xsd.exe创建xsd文件时出现错误，则使用前面提到的在msdn上的XmlSchemaInference类。下面是一个要演示的单元测试:

using System.Xml;
using System.Xml.Schema;


[TestMethod]
public void GenerateXsdFromXmlTest()
{
string folder = @"C:\mydir\mydata\xmlToCSharp";
XmlReader reader = XmlReader.Create(folder + "\some_xml.xml");
XmlSchemaSet schemaSet = new XmlSchemaSet();
XmlSchemaInference schema = new XmlSchemaInference();


schemaSet = schema.InferSchema(reader);




foreach (XmlSchema s in schemaSet.Schemas())
{
XmlWriter xsdFile = new XmlTextWriter(folder + "\some_xsd.xsd", System.Text.Encoding.UTF8);
s.Write(xsdFile);
xsdFile.Close();
}
}


// now from the visual studio command line type: xsd some_xsd.xsd /classes
小开

您只需将Cars汽车属性的一个属性从XmlArrayItem更改为xmlelement。也就是说，从

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

来

[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlElement("Car")]
public Car[] Car { get; set; }
}
小开
这个想法是让所有级别都被处理为反序列化 请参阅解决我类似问题的示例解决方案

<?xml version="1.0" ?>
<TRANSACTION_RESPONSE>
<TRANSACTION>
<TRANSACTION_ID>25429</TRANSACTION_ID>
<MERCHANT_ACC_NO>02700701354375000964</MERCHANT_ACC_NO>
<TXN_STATUS>F</TXN_STATUS>
<TXN_SIGNATURE>a16af68d4c3e2280e44bd7c2c23f2af6cb1f0e5a28c266ea741608e72b1a5e4224da5b975909cc43c53b6c0f7f1bbf0820269caa3e350dd1812484edc499b279</TXN_SIGNATURE>
<TXN_SIGNATURE2>B1684258EA112C8B5BA51F73CDA9864D1BB98E04F5A78B67A3E539BEF96CCF4D16CFF6B9E04818B50E855E0783BB075309D112CA596BDC49F9738C4BF3AA1FB4</TXN_SIGNATURE2>
<TRAN_DATE>29-09-2015 07:36:59</TRAN_DATE>
<MERCHANT_TRANID>150929093703RUDZMX4</MERCHANT_TRANID>
<RESPONSE_CODE>9967</RESPONSE_CODE>
<RESPONSE_DESC>Bank rejected transaction!</RESPONSE_DESC>
<CUSTOMER_ID>RUDZMX</CUSTOMER_ID>
<AUTH_ID />
<AUTH_DATE />
<CAPTURE_DATE />
<SALES_DATE />
<VOID_REV_DATE />
<REFUND_DATE />
<REFUND_AMOUNT>0.00</REFUND_AMOUNT>
</TRANSACTION>
</TRANSACTION_RESPONSE>

上面的XML分两层处理

  [XmlType("TRANSACTION_RESPONSE")]
public class TransactionResponse
{
[XmlElement("TRANSACTION")]
public BankQueryResponse Response { get; set; }


}

内在层面

public class BankQueryResponse
{
[XmlElement("TRANSACTION_ID")]
public string TransactionId { get; set; }


[XmlElement("MERCHANT_ACC_NO")]
public string MerchantAccNo { get; set; }


[XmlElement("TXN_SIGNATURE")]
public string TxnSignature { get; set; }


[XmlElement("TRAN_DATE")]
public DateTime TranDate { get; set; }


[XmlElement("TXN_STATUS")]
public string TxnStatus { get; set; }




[XmlElement("REFUND_DATE")]
public DateTime RefundDate { get; set; }


[XmlElement("RESPONSE_CODE")]
public string ResponseCode { get; set; }




[XmlElement("RESPONSE_DESC")]
public string ResponseDesc { get; set; }


[XmlAttribute("MERCHANT_TRANID")]
public string MerchantTranId { get; set; }


}
同样地，你需要多个级别的car as array # EYZ0 < / p >
小开

试试这个通用类的Xml序列化&反序列化。

public class SerializeConfig<T> where T : class
{
public static void Serialize(string path, T type)
{
var serializer = new XmlSerializer(type.GetType());
using (var writer = new FileStream(path, FileMode.Create))
{
serializer.Serialize(writer, type);
}
}


public static T DeSerialize(string path)
{
T type;
var serializer = new XmlSerializer(typeof(T));
using (var reader = XmlReader.Create(path))
{
type = serializer.Deserialize(reader) as T;
}
return type;
}
}
小开

我的解决方案:

  1. 使用Edit > Past Special > Paste XML As Classes获取代码中的类
  2. 尝试这样做:创建该类的列表(List<class1>)，然后使用XmlSerializer将该列表序列化到xml文件。
  3. 现在只需用数据替换文件主体，并尝试deserialize它。

代码:

StreamReader sr = new StreamReader(@"C:\Users\duongngh\Desktop\Newfolder\abc.txt");
XmlSerializer xml = new XmlSerializer(typeof(Class1[]));
var a = xml.Deserialize(sr);
sr.Close();

注意:必须注意根名称，不要更改它。我的是ArrayOfClass1

小开

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。

小开

Kevin的回答很好，但事实上，在现实世界中，您通常无法修改原始XML以满足您的需要。

对于原始XML也有一个简单的解决方案:

[XmlRoot("Cars")]
public class XmlData
{
[XmlElement("Car")]
public List<Car> Cars{ get; set; }
}


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

然后你可以简单地调用:

var ser = new XmlSerializer(typeof(XmlData));
var data = (XmlData)ser.Deserialize(XmlReader.Create(PathToCarsXml));
小开

对于初学者来说

我发现这里的答案非常有用，也就是说我仍然挣扎(只是一点点)让它工作。所以，为了帮助别人，我将详细说明工作解决方案:

来自原始问题的XML。xml位于Class1.xml文件中，代码中使用该文件的path来定位该xml文件。

我使用@erymski的答案来让它工作，所以创建了一个名为Car.cs的文件，并添加以下内容:

using System.Xml.Serialization;  // Added


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; }
}

@erymski提供的另一段代码…

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

．.． 进入你的主程序(program .cs)，在static CarCollection XCar()中，如下所示:

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


namespace ConsoleApp2
{
class Program
{


public static void Main()
{
var c = new CarCollection();


c = XCar();


foreach (var k in c.Cars)
{
Console.WriteLine(k.Make + " " + k.Model + " " + k.StockNumber);
}
c = null;
Console.ReadLine();


}
static CarCollection XCar()
{
using (TextReader reader = new StreamReader(@"C:\Users\SlowLearner\source\repos\ConsoleApp2\ConsoleApp2\Class1.xml"))
{
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
return (CarCollection)serializer.Deserialize(reader);
}
}
}
}

希望能有所帮助:-)

小开

一个衬套:

var object = (Cars)new XmlSerializer(typeof(Cars)).Deserialize(new StringReader(xmlString));

提交答案