如何反序列化 XML 文档
-
21-08-2019 - |
题
如何反序列化此 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;
}
}
这似乎不起作用:-(
解决方案
下面是一个工作版本。我因为在XML中StockNumber,品牌和型号值的元素,没有属性改变XmlElementAttribute标签的XmlElement。此外,我除去reader.ReadToEnd(); (即功能读取整个流和回报一个字符串,所以Deserialze()函数不能使用读卡器了......的位置是在流的末尾)。我还拍了几张调戏命名:)
。下面是类:
[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; }
}
在反序列化功能:
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();
和略微改动的XML(我需要添加新的元件来包装<汽车> ... Net是约反序列化阵列挑剔):
<?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>
其他提示
不如将 xml 保存到文件中,然后使用 xsd 生成 C# 类?
- 将文件写入磁盘(我将其命名为 foo.xml)
- 生成 xsd:
xsd foo.xml
- 生成 C#:
xsd foo.xsd /classes
Et voila - 和 C# 代码文件应该能够通过以下方式读取数据 XmlSerializer
:
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)
你有两种可能性。
方法一。 XSD 工具
假设您的 XML 文件位于此位置
C:\path\to\xml\file.xml
- 打开 开发者命令提示符
您可以在以下位置找到它:Start Menu > Programs > Microsoft Visual Studio 2012 > Visual Studio Tools
或者,如果您使用的是 Windows 8,则可以直接开始输入 开发者命令提示符 在 开始屏幕 - 通过键入以下内容将位置更改为 XML 文件目录
cd /D "C:\path\to\xml"
- 创造 XSD文件 从您的 xml 文件中输入
xsd file.xml
- 创造 C# 类 通过输入
xsd /c file.xsd
就是这样!您已从 xml 文件生成了 C# 类 C:\path\to\xml\file.cs
方法 2 - 选择性粘贴
需要 Visual Studio 2012+
- 将 XML 文件的内容复制到剪贴板
- 添加到您的解决方案新的空类文件(转移+替代+C)
- 打开该文件并在菜单中单击
Edit > Paste special > Paste XML As Classes
就是这样!
用法
这个辅助类的用法非常简单:
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());
}
}
}
您现在所要做的就是:
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>();
下面的代码段应达到目的(和可以忽略大部分的序列化的属性):
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);
}
请参阅如果这有助于:
[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; }
}
和如若不然使用随Visual Studio来创建一个基于XML的文件架构文档,然后再次使用它基于模式文档创建一个类的XSD.EXE程序。
我不认为.NET是“挑剔反序列化阵列”。第一个XML文档的格式不正确。 没有根元素,虽然它看起来像有。规范的XML文档有一个根和至少1个元素(如果有的话)。在您的示例:
<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>
如果您的.xml文件已经在磁盘的某处产生尝试的代码块,如果你有使用List<T>
:
//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();`
请注意:C:\serialize.xml
是我的.xml文件的路径。你可以改变它为您的需求。
凯文的雁还是不错的,除了一个事实,即在现实世界中,你往往不能够改变原始XML,以满足您的需求。
有原始XML的简单解决方案,也:
[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; }
}
然后你就可以直接电话咨询:
var ser = new XmlSerializer(typeof(XmlData));
XmlData data = (XmlData)ser.Deserialize(XmlReader.Create(PathToCarsXml));
尝试这种通用类,使用XML序列&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;
}
}
对于初学者
我发现这里的答案是非常有帮助的,那说我却还在勉强(只是一个位)来得到这个工作。因此,在情况下,它可以帮助别人,我会拼出来的工作液:
从原始的问题XML。 XML是在一个文件中Class1.xml,一个path
到本文件中的代码被用于定位该XML文件。
我使用@erymski的答案,得到这个工作,所以创建了一个名为Car.cs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; } }
的代码的其它位由@erymski ...
提供using (TextReader reader = new StreamReader(path)) { XmlSerializer serializer = new XmlSerializer(typeof(CarCollection)); return (CarCollection) serializer.Deserialize(reader); }
...进入主程序(Program.cs中),在static CarCollection XCar()
这样的:
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);
}
}
}
}
希望它能帮助: - )
我们的想法是已经全部一级正在反序列化处理 请看到解决的样品溶液我类似的问题
<?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是在两个级别进行处理
[XmlType("TRANSACTION_RESPONSE")]
public class TransactionResponse
{
[XmlElement("TRANSACTION")]
public BankQueryResponse Response { get; set; }
}
内在水平
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; }
}
相同的方式,需要car as array
多级
检查这个例子中为多电平的反序列化
如果你所提到的 .schema.xmlschemainference.aspx”相对= “nofollow”>。这里的一个单元测试以证明:
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
你可以改变从XmlArrayItem您汽车车内财物,以XmlElment一个属性。也就是说,从
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlArrayItem(typeof(Car))]
public Car[] Car { get; set; }
}
到
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlElement("Car")]
public Car[] Car { get; set; }
}
我的解决办法:
- 使用
Edit > Past Special > Paste XML As Classes
得到类代码 - 尝试这样的事情:创建一个类(
List<class1
>),然后使用XmlSerializer
序列化一个列表到xml
文件列表 。
- 现在,您只需更换你的数据文件的主体,并尝试
deserialize
它。
醇>
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();
请注意:您一定要注意根名称,不要改变它。矿是 “ArrayOfClass1”
如何一个通用类反序列化XML文档
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
// 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;
}
这部分可以或可以不是必需的。打开Visual Studio中的XML文档,右键点击XML,选择属性。然后选择你的模式文件。
async public static Task<JObject> XMLtoNETAsync(XmlDocument ToConvert)
{
//Van XML naar JSON
string jsonText = await Task.Run(() => JsonConvert.SerializeXmlNode(ToConvert));
//Van JSON naar .net object
var o = await Task.Run(() => JObject.Parse(jsonText));
return o;
}