[英]How to Deserialize XML document
我如何反序列化這個 XML 文檔:
<?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>
我有這個:
[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;
}
}
這似乎不起作用:-(
將xml 保存到文件中,然后使用xsd生成C# 類如何?
xsd foo.xml
xsd foo.xsd /classes
瞧 - 和 C# 代碼文件應該能夠通過XmlSerializer
讀取數據:
XmlSerializer ser = new XmlSerializer(typeof(Cars));
Cars cars;
using (XmlReader reader = XmlReader.Create(path))
{
cars = (Cars) ser.Deserialize(reader);
}
(在項目中包含生成的foo.cs)
這是一個工作版本。 我改變了XmlElementAttribute
標簽XmlElement
因為在StockNumber,品牌和型號值的元素,沒有屬性的XML。 我也刪除了reader.ReadToEnd();
(該函數讀取整個流並返回一個字符串,因此Deserialize()
函數不能再使用讀取器......位置在流的末尾)。 我還對命名采取了一些自由:)。
以下是課程:
[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();
和稍微調整的 xml(我需要添加一個新元素來包裝 <Cars>...Net 對反序列化數組很挑剔):
<?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>
你有兩種可能性。
C:\\path\\to\\xml\\file.xml
Start Menu > Programs > Microsoft Visual Studio 2012 > Visual Studio Tools
找到它,或者如果您使用的是 Windows 8,則可以在“開始”屏幕中開始鍵入Developer Command Promptcd /D "C:\\path\\to\\xml"
位置更改為您的 XML 文件目錄xsd file.xml
從您的 xml 文件創建XSD 文件xsd /c file.xsd
創建C# 類就是這樣! 您已經從C:\\path\\to\\xml\\file.cs
xml 文件生成了 C# 類
Edit > Paste special > Paste XML As Classes
就是這樣!
這個助手類的用法非常簡單:
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()); } } }
你現在要做的就是:
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>();
以下代碼段應該可以解決問題(您可以忽略大多數序列化屬性):
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);
}
看看這是否有幫助:
[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; }
}
如果失敗,則使用 Visual Studio 附帶的 xsd.exe 程序基於該 xml 文件創建架構文檔,然后再次使用它創建基於架構文檔的類。
我不認為 .net '對反序列化數組很挑剔'。 第一個 xml 文檔格式不正確。 沒有根元素,雖然看起來有。 規范的 xml 文檔有一個根和至少 1 個元素(如果有的話)。 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>
如果您的 .xml 文件已在磁盤某處生成並且您使用過List<T>
嘗試使用此代碼塊:
//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 文件的路徑。 您可以根據需要更改它。
Kevin 的 anser 很好,除此之外,在現實世界中,您通常無法更改原始 XML 以滿足您的需要。
原始 XML 也有一個簡單的解決方案:
[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; }
}
然后你可以簡單地調用:
var ser = new XmlSerializer(typeof(XmlData));
XmlData data = (XmlData)ser.Deserialize(XmlReader.Create(PathToCarsXml));
我發現這里的答案非常有幫助,也就是說我仍然努力(只是有點)讓這個工作。 因此,如果它對某人有幫助,我將詳細說明工作解決方案:
來自原始問題的 XML。 該 xml 位於文件 Class1.xml 中,該文件的path
用於在代碼中定位該 xml 文件。
我使用@erymski 的答案來解決這個問題,因此創建了一個名為 Car.cs 的文件並添加了以下內容:
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 提供的另一段代碼......
using (TextReader reader = new StreamReader(path)) { XmlSerializer serializer = new XmlSerializer(typeof(CarCollection)); return (CarCollection) serializer.Deserialize(reader); }
...進入您的主程序 (Program.cs),在static CarCollection XCar()
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);
}
}
}
}
希望它有幫助:-)
試試這個用於 Xml 序列化和反序列化的通用類。
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;
}
}
一個班輪:
var object = (Cars)new XmlSerializer(typeof(Cars)).Deserialize(new StringReader(xmlString));
如何使用通用類反序列化 XML 文檔
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
// 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;
}
這部分可能是必要的,也可能不是。 在 Visual Studio 中打開 XML 文檔,右鍵單擊 XML,選擇屬性。 然后選擇您的架構文件。
這個想法是處理所有級別以進行反序列化請查看解決我類似問題的示例解決方案
<?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 分兩級處理
[XmlType("TRANSACTION_RESPONSE")]
public class TransactionResponse
{
[XmlElement("TRANSACTION")]
public BankQueryResponse Response { get; set; }
}
內在層次
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; }
}
同樣的方式,您需要將car as array
多級檢查此示例以了解多級反序列化
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;
}
如果您在使用 xsd.exe 創建 xsd 文件時遇到錯誤,請使用msdn 上提到的 XmlSchemaInference 類。 這是一個單元測試來演示:
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
您只需將 Cars 汽車屬性的一個屬性從 XmlArrayItem 更改為 XmlElment。 也就是說,從
[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; }
}
我的解決方案:
Edit > Past Special > Paste XML As Classes
來獲取代碼中的類List<class1
>),然后使用XmlSerializer
將該列表序列化為xml
文件。deserialize
。代碼:
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:一定要注意根名稱,不要隨意更改。 我的是“ArrayOfClass1”
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