Read xml without path

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Read xml without path

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#1 by (1 votes)

1

What a good day, I've seen many examples of how to read xml files, but they do it from the path of the file. However, I want to search the xml files. with the OpenFileDialog (It should be noted that the xml have the same structure, the nodes do not change only the data). This is my example of my xml file:

Now I will show you my code what I am trying to do:

  private void btn3_Click(object sender, EventArgs e)
    {
        Stream myStream = null;
        OpenFileDialog OpenFileDialog1 = new OpenFileDialog();



        OpenFileDialog1.InitialDirectory = "C:\Users\Prueba";
        OpenFileDialog1.Filter = "xml files (*.xml)|*.xml|Todos los archivos (*.*)|*.*";
        OpenFileDialog1.FilterIndex = 2;
        OpenFileDialog1.RestoreDirectory = true;
        OpenFileDialog1.Multiselect = true;


        if (OpenFileDialog1.ShowDialog() == DialogResult.OK)
        {
            try
            {

                if ((myStream = OpenFileDialog1.OpenFile()) != null)
                {
                    XmlTextReader xmlTextReader = new XmlTextReader(""); //Aquí pondría la ruta de origen del archivo pero eso no es lo que yo quiero.
                    string etiquetafinal = "";
                    while (xmlTextReader.Read())
                    {
                        if (xmlTextReader.NodeType == XmlNodeType.Element)
                        {
                            rtxt1.Text += (new string(' ', xmlTextReader.Depth * 3) + "<" + xmlTextReader.Name + ">");
                            etiquetafinal = xmlTextReader.Name;
                            continue;
                        }
                        if (xmlTextReader.NodeType == XmlNodeType.Text)
                        {
                            rtxt1.Text += xmlTextReader.ReadContentAsString() + "</" + etiquetafinal + ">";

                        }
                        else
                        {
                            rtxt1.Text += "\r";
                        }
                    }
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
            }
             }


    }

I hope you understand me, I want to find the xml file and read it. thanks.

c# windows-forms xml

asked by SoyKrut 04.12.2017 в 22:31

source

1 answer

position back and above Rename the package of my project in Android Studio

score 1 · Accepted Answer

OpenFileDialog has a property called FileName where at the time of selecting the file, you will have the path of it:

if ((myStream = OpenFileDialog1.OpenFile()) != null)
{
    // le envias el filename
    XmlTextReader xmlTextReader = new XmlTextReader(OpenFileDialog1.FileName); 
//...

In case the user can select more than 1 file, then you can use FileNames , which would be a string array of the selected files:

if ((myStream = OpenFileDialog1.OpenFile()) != null)
{

    foreach(string filename OpenFileDialog1.FileNames)
    {

        // le envias el filename
        XmlTextReader xmlTextReader = new XmlTextReader(filename); 
        //...