Add a xml node through xslt or C# code

Question

I have an input xml as follows,

<?xml version="1.0" encoding="utf-8" ?>
<?xml-stylesheet type="text/xsl" href="cdcatalog.xsl"?>
<root>
  <employee>
    <firstname>Kaushal</firstname>
    <lastname>Parik</lastname>
  </employee>
  <employee>
    <firstname>Abhishek</firstname>
    <lastname>Swarnkar</lastname>
  </employee>
</root>

and I need output xml as

<?xml version="1.0" encoding="utf-8" ?>
<?xml-stylesheet type="text/xsl" href="cdcatalog.xsl"?>
<root>
  <employee>
    <firstname>Kaushal</firstname>
    <lastname>Parik</lastname>
    <status>Single</status>
  </employee>
  <employee>
    <firstname>Abhishek</firstname>
    <lastname>Swarnkar</lastname>
    <status>Single</status>
  </employee>
</root>

The value of "status" is "Single" in all the nodes.... I know how to add this static text "Single" through c# code.... But, I don't know how to add the node "status" in xml through xslt.... When I try, it gets added below the node "firstname" and not in the expected place as shown.... Please help me how can i achieve this.... The xslt and C# code used by me are,

XSLT:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
            xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"
            xmlns:myUtils="pda:MyUtils">
  <xsl:output method="xml" indent="yes"/>
  <xsl:template match="@*|node()">
    <xsl:copy>
      <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="employee/firstname">
    <xsl:element name="firstname">
      <xsl:value-of select="myUtils:FormatName(.)" />
    </xsl:element>
    <xsl:element name ="status">
      <xsl:value-of select ="Single"/>
    </xsl:element>
  </xsl:template>
</xsl:stylesheet>

aspx.cs:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Xml;
using System.Xml.Xsl;
using System.Xml.XPath;
using System.IO;

public partial class nirav : System.Web.UI.Page
{
    public class MyXslExtension
    {
        public string FormatName(string name)
        {
            return "Mr. " + name;
        }
        public int GetAge(string name)
        {
            int age = name.Count();
            return age;
        }
    }  
    protected void Page_Load(object sender, EventArgs e)
    {
        string outputpath = "nirav.xml";
        XsltArgumentList arguments = new XsltArgumentList();
        arguments.AddExtensionObject("pda:MyUtils", new MyXslExtension());
        using (StreamWriter writer = new StreamWriter(outputpath))
        {
            XslCompiledTransform transform = new XslCompiledTransform();
            transform.Load("http://localhost:4329/XsltTransform/nirav.xslt");
            transform.Transform("http://localhost:4329/XsltTransform/nirav.xml", arguments, writer);
        }

    }
}

Your help is greatly appreciated....

Do you definitely want to use XSLT for this? It looks like a really simple transformation which code be done very easily in code. — Jon Skeet
– Jon Skeet, Commented Aug 3, 2012 at 7:38
@Nirav: Yes, that's better - although I think it would be worth removing the "Please mark ACCEPT or +1 if it is useful to you...." from all your answers. (It's really unnecessary, and looks like you're just desperate for rep...) — Jon Skeet
– Jon Skeet, Commented Aug 3, 2012 at 7:47
@JonSkeet: I definitely need to use xslt.... This is not really my project.... I am just giving a sample similar to my project in order to give a clear picture of how my code should work.... Thanks anyway.... — DotNet Beginner
– DotNet Beginner, Commented Aug 3, 2012 at 9:19
@NivethanRajendran: Okay - then my answer won't be useful to you, but I'll leave it up in case it helps future readers. — Jon Skeet
– Jon Skeet, Commented Aug 3, 2012 at 9:19

Tim C · Accepted Answer · 2012-08-03 07:42:13Z

There are a couple of issues with your XSLT. Firstly, this expression is incorrect

<xsl:value-of select="Single"/>

This will select the value of the element Single, which does not exist in your input XML. You actually want to output the literal value 'Single'

<xsl:value-of select="'Single'"/>

Or rather, you could just output the whole element 'as-is'

<status>Single</status>

Secondly, it looks like you want to want to add the status as the last element of the employee element. In which case, you need a template to match the employee element, which copies all existing elements, and then just adds the new status element

<xsl:template match="employee">
   <xsl:copy>
      <xsl:apply-templates select="@*|node()"/>
      <status>Single</status>
   </xsl:copy>
</xsl:template>

Here is the full XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl">
   <xsl:output method="xml" indent="yes"/>

   <xsl:template match="@*|node()">
      <xsl:copy>
         <xsl:apply-templates select="@*|node()"/>
      </xsl:copy>
   </xsl:template>

   <xsl:template match="employee">
      <xsl:copy>
         <xsl:apply-templates select="@*|node()"/>
         <status>Single</status>
      </xsl:copy>
   </xsl:template>
</xsl:stylesheet>

When applied to your XML, the following is output

<root>
   <employee>
      <firstname>Kaushal</firstname>
      <lastname>Parik</lastname>
      <status>Single</status>
   </employee>
   <employee>
      <firstname>Abhishek</firstname>
      <lastname>Swarnkar</lastname>
      <status>Single</status>
   </employee>
</root>

(Note, I've removed the reference to the extension functions, because I don't have those myself on my PC).

Jon Skeet · Accepted Answer · 2012-08-03 07:38:59Z

2

Personally I think using XSLT is overkill for this. I would just use:

XDocument doc = XDocument.Load("http://localhost:4329/XsltTransform/nirav.xml");
foreach (var employee in doc.Descendants("employee"))
{
    employee.Add(new XElement("status", "Single"));
}
doc.Save(outputPath);

Of course if you have other reasons for using XSLT, that's fine - just don't think it's the only way of modifying XML in .NET :)

answered Aug 3, 2012 at 7:38

Jon Skeet

1.5m893 gold badges9.3k silver badges9.3k bronze badges

Comments

user854301 · Accepted Answer · 2012-08-03 07:44:41Z

1

You can do with linq to xml:

    var document = XDocument.Parse(xml);

    foreach (var element in document.Root.Elements("employee"))
    {
        element.Add(new XElement("status", "Single"));
    }

edited Aug 3, 2012 at 7:44

answered Aug 3, 2012 at 7:38

user854301

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Comments

Suspitsyn Kirill · Accepted Answer · 2012-08-03 07:36:59Z

0

How about

<root>
    <xsl:for-each select="\\root\employee">
        <employee>
            <xsl:copy-of select="firstname"/>
            <xsl:copy-of select="lastname"/>
            <status>Single</status>
        </employee>
    </xsl:for-each>
</root>

answered Aug 3, 2012 at 7:36

Suspitsyn Kirill

344 bronze badges

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