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How to open a .txt file using openFileDialog in C #?

I need to open and read from a file .txt, here is the code I use:

Stream myStream;
openFileDialog1.FileName = string.Empty; 
openFileDialog1.InitialDirectory = "F:\\";
if (openFileDialog1.ShowDialog() == DialogResult.OK) 
{
    var compareType = StringComparison.InvariantCultureIgnoreCase;
    var fileName = Path.GetFileNameWithoutExtension(openFileDialog1.FileName);
    var extension = Path.GetExtension(openFileDialog1.FileName);
    if (extension.Equals(".txt", compareType))
    {
        try 
        { 
            using (myStream = openFileDialog1.OpenFile()) 
            { 
                string file = Path.GetFileName(openFileDialog1.FileName);
                string path = Path.GetFullPath(file); //when i did it like this it work fine but all the time give me same path whatever where my "*.txt" file is
                //Insert code to read the stream here. 
                //fileName = openFileDialog1.FileName; 
                StreamReader reader = new StreamReader(path);
                MessageBox.Show(file, "fileName");
                MessageBox.Show(path, "Directory");
            } 
        } 
        // Exception thrown: Empty path name is not legal
        catch (ArgumentException ex) 
        { 
            MessageBox.Show("Error: Could not read file from disk. " +
                            "Original error: " + ex.Message); 
        } 
    }
    else 
    {
        MessageBox.Show("Invaild File Type Selected");
    } 
}

The above code throws an exception that says: "An empty path is not legal."

What am I doing wrong?

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4 answers

As hmemcpy pointed out , your problem is the following lines

using (myStream = openFileDialog1.OpenFile())
{
   string file = Path.GetFileName(openFileDialog1.FileName);
   string path = Path.GetDirectoryName(file);
   StreamReader reader = new StreamReader(path);
   MessageBox.Show(file, "fileName");
   MessageBox.Show(path, "Directory");
}

I'm going to break for you:

/*
 * Opend the file selected by the user (for instance, 'C:\user\someFile.txt'), 
 * creating a FileStream
 */
using (myStream = openFileDialog1.OpenFile())
{
   /*
    * Gets the name of the the selected by the user: 'someFile.txt'
    */
   string file = Path.GetFileName(openFileDialog1.FileName);

   /*
    * Gets the path of the above file: ''
    *
    * That because the above line gets the name of the file without any path.
    * If there is no path, there is nothing for the line below to return
    */
   string path = Path.GetDirectoryName(file);

   /*
    * Try to open a reader for the above bar: Exception!
    */
   StreamReader reader = new StreamReader(path);

   MessageBox.Show(file, "fileName");
   MessageBox.Show(path, "Directory");
}

What you have to do is build the code into something like

using (myStream = openFileDialog1.OpenFile())
{
   // ...
   var reader = new StreamReader(myStream);
   // ...
}
+7
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Do you want the user to select only .txt files? Then use the .Filter property , for example:

openFileDialog1.Filter = "txt files (*.txt)|*.txt";
+6
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:

string file = Path.GetFileName(openFileDialog1.FileName);
string path = Path.GetDirectoryName(file);

file , . MyFile.txt, path. StreamReader , .

By the way, this is exactly what the exception tells you. If you delete try..catcharound the usage block, you would see how this happened during debugging in your Visual Studio.

+5
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StreamReader takes a Stream type when you pass it a string. try it,

Stream myStream;

        using (myStream = openFileDialog1.OpenFile())
        {
            string file = Path.GetFileName(openFileDialog1.FileName);
            string file2 = Path.GetFileNameWithoutExtension(openFileDialog1.FileName);

            string path = Path.GetDirectoryName(openFileDialog1.FileName);

            StreamReader reader = new StreamReader(myStream);

            while (!reader.EndOfStream)
            {
                MessageBox.Show(reader.ReadLine());
            }

            MessageBox.Show(openFileDialog1.FileName.ToString());
            MessageBox.Show(file, "fileName");
            MessageBox.Show(path, "Directory");
        }
+1
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