如何使用C＃更新xml文件

Question

I have a requirement to update xml file by reading another xml file.

eg:

Source.xml

<?xml version="1.0" encoding="utf-8"?>
<packages>
  <package id="package1" version="1.0.26.0"/>
  <package id="package3" version="1.0.12.0"/>
  <package id="package4" version="1.0.40.0"/>
  <package id="package12" version="1.0.38.0"/>
  <package id="package6" version="1.0.8.0"/>
</packages>

Target.xml

<?xml version="1.0" encoding="utf-8"?>
<packages>
  <package id="package1" version="1.0.2.0"/>
  <package id="package2" version="1.0.1.0"/>
  <package id="package3" version="1.0.4.0"/>
  <package id="package4" version="1.0.3.0"/>
</packages>

In the above example, "package1" exists in both the files. So the version of "package1" in Target.xml has to get update with the value in Source.xml.

I tried the below code, but encountered exception during execution(no error during compilation):

Unhandled Exception: System.NullReferenceException: Object reference not set to an instance of an object.

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

namespace XMLUpdate
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine("In Execute..");
            string SourceFile = args[0];
            string TargetFile = args[1];
            string SourcePackageId = null;
            string SourcePackageVersion = null;
            XmlDocument SourceXmlDoc = new XmlDocument();
            XmlDocument TargetXmlDoc = new XmlDocument();
            SourceXmlDoc.Load(SourceFile);
            TargetXmlDoc.Load(TargetFile);
            XmlElement SourceRootElement = SourceXmlDoc.DocumentElement;
            //XmlElement SourceElement = SourceRootElement["packages"];
            XmlNodeList SourcexnList = SourceXmlDoc.SelectNodes("/packages");
            foreach (XmlNode Sourcexn in SourcexnList)
            {
                Console.WriteLine("In source loop..");
                SourcePackageId = Sourcexn["id"].InnerText;
                SourcePackageVersion = Sourcexn["version"].InnerText;
                Console.WriteLine("In source loop SourcePackageId.." + SourcePackageId);
                XmlElement TargetRootElement = TargetXmlDoc.DocumentElement;
                XmlNodeList TargetxnList = TargetXmlDoc.SelectNodes("/packages");
                foreach (XmlNode Targetxn in TargetxnList)
                {
                    Console.WriteLine("In Target loop..");
                    string TargetPackageId = Targetxn["id"].InnerText;
                    //string TargetPackageVersion = Targetxn["version"].InnerText;

                    if (SourcePackageId.Equals(TargetPackageId))
                    {
                        Targetxn["version"].InnerText = SourcePackageVersion;
                    }
                }
            }
        }
    }
}

Any help?

Thanks

Answer 1

Try the following:

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

namespace XMLUpdate
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine("In Execute..");
            string SourceFile = args[0];
            string TargetFile = args[1];
            string SourcePackageId = null;
            string SourcePackageVersion = null;
            XmlDocument SourceXmlDoc = new XmlDocument();
            XmlDocument TargetXmlDoc = new XmlDocument();
            SourceXmlDoc.Load(SourceFile);
            TargetXmlDoc.Load(TargetFile);
            XmlElement SourceRootElement = SourceXmlDoc.DocumentElement;
            //XmlElement SourceElement = SourceRootElement["packages"];
            XmlNodeList SourcexnList = SourceXmlDoc.SelectNodes("/packages");
            foreach (XmlNode Sourcexn in SourcexnList)
            {
                Console.WriteLine("In source loop..");
                foreach(XmlNode childS in Sourcexn.ChildNodes)
                {
                    SourcePackageId = childS.Attributes["id"].InnerText;
                    SourcePackageVersion = childS.Attributes["version"].InnerText;
                    Console.WriteLine("In source loop SourcePackageId.." + SourcePackageId);
                    XmlElement TargetRootElement = TargetXmlDoc.DocumentElement;
                    XmlNodeList TargetxnList = TargetXmlDoc.SelectNodes("/packages");
                    foreach (XmlNode Targetxn in TargetxnList)
                    {
                        Console.WriteLine("In Target loop..");
                        foreach (XmlNode childT in Targetxn.ChildNodes)
                        {
                            string TargetPackageId = childT.Attributes["id"].InnerText;
                            if (SourcePackageId.Equals(TargetPackageId))
                            {
                                childT.Attributes["version"].InnerText = SourcePackageVersion;
                            }
                        }
                    }
                }
            }
        }


    }
}

You need to check inside the ChildNodes of the xmlNode.

Answer 2

This should work, just include System.Xml.Linq and System.Linq .

var source = XDocument.Load("Source.xml");
var sPackages = source.Descendants("package");

var target = XDocument.Load("Target.xml");
var tPackages = target.Descendants("package");

foreach (var sPackage in sPackages)
{
    var tPackage = tPackages.FirstOrDefault(x => x.Attribute("id").Value == sPackage.Attribute("id").Value);

    if (tPackage != null && tPackage.Attribute("version").Value != sPackage.Attribute("version").Value)
    {
        target.Descendants("package").Single(x => x.Attribute("id").Value == sPackage.Attribute("id").Value)
                                     .SetAttributeValue("version", sPackage.Attribute("version").Value);
    }
}

target.Save("Target.xml");

This will update every target file's package version number that is also in the source file. You can easily add more functionality based on the above code. As you see, Linq to XML is very clean and easy.