How to create a XML file using XDocument and LINQ in C#?

Question

I have a three List in c# ,the variable names are l_lstData1,l_lstData2,l_lstData3

The File Structure is

 <FileDetails>
 <Date FileModified="29/04/2010 12:34:02" /> 
 <Data Name="Data_1" DataList="India" Level="2" /> 
 <Data Name="Data_2" DataList="chennai" Level="2" /> 
 <Data Name="Data_3" DataList="hyderabad" Level="2" /> 
 <Data Name="Data_4" DataList="calcutta" Level="2" /> 
 <Data Name="Data_5" DataList="vijayawada" Level="1" /> 
 <Data Name="Data_6" DataList="cochin" Level="1" /> 
 <Data Name="Data_7" DataList="madurai" Level="0" /> 
 <Data Name="Data_8" DataList="trichy" Level="0" /> 
 </FileDetails>

The Values od 3 Lists are as follows :

  l_lstData1[0] = "India";l_lstData1[1] = "chennai";l_lstData1[2] = "hyderabad";
  l_lstData1[3] = "calcutta";

so the level attribute of the above XML(element : Data) has tha value = "2".

  l_lstData2[0] = "vijayawada";l_lstData2[1] = "cochin";

so the level attribute of the above XML(element : Data) has tha value = "1".

 l_lstData3[0] = "madurai";l_lstData3[1] = "trichy";

so the level attribute of the above XML(element : Data) has tha value = "0".

How can i create the XML using Xdocument and also using LINQ....Plz revert back me if u have any queries

You say: so the level attribute of the above XML(element : Data) has tha value = "2" (and likewise for the other values) - but I don't understand why that makes it "so". Just because of which list it's in? Or because of something else? — Jon Skeet
– Jon Skeet, Commented Jun 25, 2010 at 5:26

Jon Skeet · Accepted Answer · 2010-06-25 05:30:35Z

2

Here's an alternative to Pramodh's solution, if I've understood it correctly:

// First build up a single list to work with, using an anonymous type
var singleList = l_lstData1.Select(x => new { Value = x, Level = 2})
         .Concat(l_lstData2.Select(x => new { Value = x, Level = 1})
         .Concat(l_lstData3.Select(x => new { Value = x, Level = 0});

var doc = new XDocument(
    new XElement("FileDetails",
        new XElement("Date",new XAttribute("FileModified", DateTime.Now)),
        singleList.Select((item, index) => new XElement("Data",
            new XAttribute("Name", "Data_" + (index + 1)),
            new XAttribute("DataList", item.Value),
            new XAttribute("Level", item.Level))));

answered Jun 25, 2010 at 5:30

Jon Skeet

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1 Comment

user374191 Over a year ago

skeet : its quite simple.Thank U

Thorin Oakenshield · Accepted Answer · 2010-06-25 05:25:26Z

Try like this:

           XDocument TEMP = new XDocument(new XElement("FileDetails",
                                          new XElement("Date",new XAttribute("FileModified", DateTime.Now.ToString())),
                   l_lstData1.Select(l => new XElement("Data",new XAttribute("Name","Data_"+(l_lstData1.IndexOf(l)+1).ToString()),
                                                              new XAttribute ("DataList",l.ToString()),
                                                              new XAttribute ("Level","Level2"))),
                   l_lstData2.Select(l => new XElement("Data",new XAttribute("Name","Data_"+(l_lstData2.Count + l_lstData2.IndexOf(l)+1).ToString()),
                                                              new XAttribute ("DataList",l.ToString()),
                                                              new XAttribute ("Level","Level1"))) ,
                   l_lstData3.Select(l => new XElement("Data",new XAttribute("Name", "Data_" + (l_lstData3.Count + l_lstData2.Count + l_lstData3.IndexOf(l) + 1).ToString()),
                                                              new XAttribute ("DataList",l.ToString()),
                                                              new XAttribute ("Level","Level0")))                                                                 

                                                        ));


        TEMP.Save("TEMP.xml");

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