I am creating a program that creates, writes, and saves an xml file. When I try to open the saved file I get an error that says the file cannot be accessed because it is being used by another process. I assumed it was because I didn't close the filestream after I saved the file so I made the correction. I still can't open the file and I receive the same error. I'm not sure what the issue is beyond this point. How can I fix this?
namespace XML_DataSets
{
public partial class FormAddNew : Form
{
XmlSerializer xs;
List<Class1> ls;
//create the DataTable
DataTable dt = new DataTable("Contact");
XDocument xd = new XDocument();
public FormAddNew()
{
InitializeComponent();
ls = new List<Class1>();
xs = new XmlSerializer(typeof(List<Class1>));
//create columns for the DataTable
DataColumn dc1 = new DataColumn("Id");
dc1.DataType = System.Type.GetType("System.Int32");
dc1.AutoIncrement = true;
dc1.AutoIncrementSeed = 1;
dc1.AutoIncrementStep = 1;
//add columns to the DataTable
dt.Columns.Add(dc1);
dt.Columns.Add(new DataColumn("Name"));
dt.Columns.Add(new DataColumn("Age"));
dt.Columns.Add(new DataColumn("Gender"));
//create DataSet
DataSet ds = new DataSet();
ds.DataSetName = "AddressBook";
ds.Tables.Add(dt);
}
private void buttonCreate_Click(object sender, EventArgs e)
{
DataRow row = dt.NewRow();
row["Name"] = textBoxName.Text;
row["Age"] = textBoxAge.Text;
row["Gender"] = textBoxGender.Text;
dt.Rows.Add(row);
dataGridView1.DataSource = dt;
//dt.WriteXml("Contacts.xml");
xd = WriteDt2Xml(dt);
}
public static XDocument WriteDt2Xml(DataTable dt)
{
using (var stream = new MemoryStream())
{
dt.WriteXml(stream);
stream.Position = 0;
XmlReaderSettings settings = new XmlReaderSettings();
settings.ConformanceLevel = ConformanceLevel.Fragment;
XmlReader reader = XmlReader.Create(stream, settings);
reader.MoveToContent();
if (reader.IsEmptyElement) { reader.Read(); return null; }
return XDocument.Load(reader);
}
}
private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
Stream input = null;
OpenFileDialog dialog = new OpenFileDialog();
openFileDialog.Filter = "xml file | *.xml";
openFileDialog.FilterIndex = 2;
openFileDialog.RestoreDirectory = true;
if (openFileDialog.ShowDialog() == DialogResult.OK)
{
try
{
if ((input = openFileDialog.OpenFile()) != null)
{
FileStream fs = new FileStream(@openFileDialog.FileName.ToString(), FileMode.Open, FileAccess.ReadWrite, FileShare.ReadWrite);
ls = (List<Class1>)xs.Deserialize(fs);
dataGridView1.DataSource = ls;
fs.Close();
}
}
catch (Exception ex)
{
MessageBox.Show(ex.Message, "ERROR");
}
}
}
}
}
@Daniel Advise taken well...I refactored the code and see the error you referred to. I checked out the two links you provided as examples. I made corrections but I still get the same result.