CODEWITHSUNDEEP
c#linqlinq-to-xml
user2729272
user2729272

Reputation: 359

LINQ to add Xattribute using loop

Here is my sample XML,

 <A>
    <B  id="ABC">
      <C name="A" />
      <C name="B" />
      <C name="C" />
      <C name="G" />
    </B>
    <B id="ZYZ">
      <C name="A" />
      <C name="B" />
      <C name="C" />
      <C name="D" />
    </B>
  </A>

I need to loop each of the <B> nodes under parent <A> and add a sequence numbered Xattribute called Sno to the <C> tag for each <B> as shown below,

 <A>
    <B  id="ABC">
      <C name="A" Sno ="1" />
      <C name="B" Sno ="2"/>
      <C name="C" Sno ="3"/>
      <C name="G" Sno ="4"/>
    </B>
    <B id="ZYZ">
      <C name="A" Sno ="1"/>
      <C name="B" Sno ="2"/>
      <C name="C" Sno ="3"/>
      <C name="D" Sno ="4"/>
    </B>
  </A>

Using following c# code,

var final = from x in afterGrouping.Descendants("A").Descendants("B").Select((sc, i) => new {sc, sequence = i + 1})
                                select new XElement("C",
                                                    new XAttribute("id", x.sc.Element("C").Attribute("id").Value),
                                                    new XAttribute("sequence", x.sequence));

Upvotes: 1

Views: 456

Answers (2)

Douglas
Douglas

Reputation: 54877

It might be easier to use a loop to alter the current document, rather than project one anew:

foreach (XElement b in xDoc.Descendants("B"))
{
    int seq = 1;
    foreach (XElement c in b.Elements("C"))
        c.Add(new XAttribute("Sno", seq++));
}

Upvotes: 4

JMK
JMK

Reputation: 28059

Something like this should do the trick:

using System.Linq;
using System.Xml.Linq;

namespace ConsoleApplication19
{
    class Program
    {
        static void Main(string[] args)
        {
            const string xml = @"<A><B  id='ABC'><C name='A' /><C name='B' /><C name='C' /><C name='G' /></B><B id='ZYZ'><C name='A' /><C name='B' /><C name='C' /><C name='D' /></B></A>";
            var doc = XDocument.Parse(xml);

            var bs = doc.Descendants("B");

            foreach (var r in bs)
            {
                var cs = r.Descendants("C");

                var xElements = cs as XElement[] ?? cs.ToArray();
                for (var i = 1; i <= xElements.Count(); i++)
                {
                    var c = xElements.ElementAt(i-1);
                    c.SetAttributeValue("Sno", i);
                }
            }

            var resultingXml = doc.ToString();
        }
    }
}

Upvotes: 0

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