Question

这里我有一个动态创建xml文件的场景，它应该是可序列化的。 xml就像：

<person>
    <personaldata>
        <name>gopi</name>
        <lastname>ch</lastname>
    </personaladata>
    <Educationaladata>
        <Graduation>b.tech</graduation>
        <designation>Engineer</designation>
    </educationaldata>
</person>

人类具有姓名，姓氏，名称，毕业和财产

我试过这个

public string CreateXmlObject(Person objPerson)    
{
    var objXmlDocument = new XmlDocument();
    var objXpath = objXmlDocument.CreateNavigator();
    var objXmlSeialiser = new XmlSerializer(objPerson.GetType());
    using (var xs = objXpath.AppendChild())
    {
        objXmlSeialiser.Serialize(xs, objPerson);
    }
    return objXmlDocument.OuterXml;
}

我的问题是我必须从Xml读取特定数据并将特定数据更新到Xml。我想在更新时只阅读Personaldata，更新应仅适用于Personaldata不Otherdata

Answer 1

所有XML的Fisrt都无效，所有节点都应与其结束标记匹配。考虑到您的Intent类看起来像这样： -

person

您可以使用public class Person { public string name { get; set; } public string lastname { get; set; } public string Graduation { get; set; } public string designation { get; set; } }轻松完成： -

LINQ-to-XML

Answer 2

试试xml linq

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            string name = "gopi";
            string lastname = "ch";
            string graduation = "b.tech";
            string designation = "Engineer";

            XElement personalData = new XElement("person", new XElement[] {
                new XElement("personaldata", new XElement[] {
                    new XElement("name", name),
                    new XElement("lastname", lastname)
                }),
                new XElement("Educationadata", new XElement[] {
                    new XElement("Graduation", graduation),
                    new XElement("designation", designation)
                })
            });

        }
    }
}

Answer 3

sSometing like -

var xml = XDocument.Load("xml path");
var personaldata = xml.Descendents("personaldata").FirstOrDefault();
if (data != null)
{
    foreach (var t in data.Descendants())
    {
        t.Value = "test";
    }                
}

从对象动态创建Xml

3 个答案: