Question

我的问题如下。

我有一个由命名空间版本化的xml。我想将这个收到的xml序列化到相应的对象中，但我知道这样做的唯一方法就是我必须处理xml两次。首先发现命名空间，然后根据发现的命名空间序列化到正确类型的对象。这对我来说似乎非常低效，并且必须有某种方式使用泛型或其他东西来获取适当类型的对象，而不使用'if namespace == x然后序列化到'检查。

以下是我知道完成此任务的唯一方法示例。是否有更好或更有效的方式？

由于

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System.Xml.Linq;
using System.Xml;
using System.Xml.Serialization;
using System.IO;

namespace TestProject1
{
    [TestClass]
    public class UnitTest1
    {
        [TestMethod]
        public void TestMethod3()
        {
            //Build up an employee object to xml
            Schema.v2.Employee employee = new Schema.v2.Employee { FirstName = "First", LastName = "Last" };
            string xml = employee.ObjectToXml<Schema.v2.Employee>();

            //Now pretend I don't know what type I am receiving.
            string nameSpace = GetNamespace(xml);
            Object newemp;
            if (nameSpace == "Employee.v2")
                newemp = XmlSerializationExtension.XmlToObject<Schema.v2.Employee>(null, xml);
            else
                newemp = XmlSerializationExtension.XmlToObject<Schema.v1.Employee>(null, xml);

            // Check to make sure that the type I got was what I made.
            Assert.AreEqual(typeof(Schema.v2.Employee), newemp.GetType());
        }

        public string GetNamespace(string s)
        {
            XDocument z = XDocument.Parse(s);
            var result = z.Root.Attributes().
                    Where(a => a.IsNamespaceDeclaration).
                    GroupBy(a => a.Name.Namespace == XNamespace.None ? String.Empty : a.Name.LocalName,
                            a => XNamespace.Get(a.Value)).
                    ToDictionary(g => g.Key,
                                 g => g.First());

            foreach (System.Xml.Linq.XNamespace item in result.Values)
                if (item.NamespaceName.Contains("Employee")) return item.NamespaceName;

            return String.Empty;
        }
    }

    public static class XmlSerializationExtension
    {
        public static string ObjectToXml<T>(this T Object)
        {
            XmlSerializer s = new XmlSerializer(Object.GetType());
            using (StringWriter writer = new StringWriter())
            {
                s.Serialize(writer, Object);
                return writer.ToString();
            }
        }
        public static T XmlToObject<T>(this T Object, string xml)
        {
            XmlSerializer s = new XmlSerializer(typeof(T));
            using (StringReader reader = new StringReader(xml))
            {
                object obj = s.Deserialize(reader);
                return (T)obj;
            }
        }
    } 
}

namespace Schema.v1
{
    [XmlRoot(ElementName = "Employee", Namespace= "Employee.v1", IsNullable = false)]
    public class Employee
    {
        [XmlElement(ElementName = "FirstName")]
        public string FirstName { get; set; }
        [XmlElement(ElementName = "LastName")]
        public string LastName { get; set; }
    }
}
namespace Schema.v2
{
    [XmlRoot(ElementName = "Employee", Namespace = "Employee.v2", IsNullable = false)]
    public class Employee
    {
        [XmlAttribute(AttributeName = "FirstName")]
        public string FirstName { get; set; }
        [XmlAttribute(AttributeName = "LastName")]
        public string LastName { get; set; }
    }
}

Answer 1

两个建议：

首先，也许根本不这样做。如果要进行序列化，除非调用者指定模式，否则首选一种方法而不是另一种方法。

其次，不要解析XML以进行发现;只需在文件的前100个字节内（或者说，你需要去获取信息）进行“Employee.v2”和“Employee.v1”的字符串匹配。这应该有效，除非这些将成为您数据中的常见字符串。

在命名空间版本化时，将xml序列化为适当的对象的最佳方法是什么？

1 个答案: