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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
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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 30

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 30

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 30

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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