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Creating XML file in Winforms to Load into a DataGrid in C# part 1

Posted on 2009-03-30
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Last Modified: 2012-05-06
I need to write a function in C# that checks if an XML file exists at a certain path. If the file exists I need to append child nodes(person) to the XML file and save it.  If it doesn't exists, I need to create The XML file , then append the child nodes(person) to this and save it.  I will later load the XML file using another form that contains a DataGrid so the file can be viewed in a grid style layout.  How would I write a function to do this? The part 2 of this question I will put in a separate question as far as loading the DataGrid. After a solution is found to this one. Thanks for any help in advance.
<Family> // root 

<person>

<firstname>Danny</firstname>

<lastname>Glover</lastname>

<address> 542 bell south</address>

</person>

<person>

<firstname>Brad</firstname>

<lastname>Smith</lastname>

<address> 508 bell south</address>

</person>

</Family>

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Question by:encoredatas
  • 3
5 Comments
 
LVL 12

Expert Comment

by:williamcampbell
ID: 24024558
Something like...          


            XmlDataDocument doc;

            doc = new XmlDataDocument();

           bool exists = true;

            try

            {

                doc.Load("C:/Work//test12.xml");

            }

            catch (System.IO.FileNotFoundException fe)

            {

                // File not found

                exists = false;

                doc.CreateElement( "Family" );

            }

            

            XmlDocument doc2 = new XmlDocument();

            doc2.LoadXml( "<person......");
 

            doc.AddChild ( doc2.DocumentElement );
 

           doc.Save ();

            

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LVL 29

Accepted Solution

by:
anarki_jimbel earned 500 total points
ID: 24025584
Try this:
            string filename = "family.xml";

            bool exists = File.Exists(filename);

            XmlDocument doc = new XmlDocument();

            if (exists)

            {

                doc.Load(filename);
 

            }

            else

            {

                XmlElement family = doc.CreateElement("family");

                doc.AppendChild(family);

            }
 

                XmlElement person = doc.CreateElement("person");

                XmlElement firstname = doc.CreateElement("firstname");

                XmlElement lastname = doc.CreateElement("lastname");

                XmlElement address = doc.CreateElement("address");

                doc.DocumentElement.AppendChild(person);

                person.AppendChild(firstname);

                person.AppendChild(lastname);

                person.AppendChild(address);

                firstname.InnerText = "Elvis";

                lastname.InnerText = "Presley";

                address.InnerText = "Hollywood";

                doc.Save(filename);

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LVL 29

Expert Comment

by:anarki_jimbel
ID: 24025593
I tested the code and it works. However it does not validates the xml. If xml is not valid - we'll get an error. So some additional precautions should be taken. E.g. put doc.Load(filename)  into try-catch
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LVL 29

Expert Comment

by:anarki_jimbel
ID: 24025725
Bonus :)

Really, I probably can give you a solution to part 2. The easiest way to display your xml in a grid is to use a dataset. If you want to just display (not edit) - set e.g. datagridview1.editmode = editprogrammatically
        private void button1_Click(object sender, EventArgs e)

        {

            string filename = "family.xml";

            DataSet ds = new DataSet();

            ds.ReadXml(filename);

            this.dataGridView1.DataSource = ds.Tables[0];

        }

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Author Closing Comment

by:encoredatas
ID: 31564600
Thanks so much this work like a charm and saved me a lot of time. P. S Thanks for the bonus.
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