如何在 c# 中使用 openFileDialog 打开 file.txt？

Question

I have to open and read from a .txt file, here is the code I'm using:

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's 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 code above throws an exception which says "Empty path name is not legal" .

What am I doing wrong?

Answer 1

As pointed by hmemcpy , your problem is in 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 down 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's 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 should do is to cahnge the code to something like

using (myStream = openFileDialog1.OpenFile())
{
   // ...
   var reader = new StreamReader(myStream);
   // ...
}

Answer 2

You want user to select only.txt files? Then use the .Filter property, like that:

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

Answer 3

Your bug is in the lines:

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

In the first line the file variable will only contain the file name, eg MyFile.txt , making the second line return an empty string to the path variable. Further down your code you'll attempt to create a StreamReader with an empty path, and this what throws the exception.

By the way, this is exactly what the exception tells you. If you remove the try..catch around the using block, you would've seen it happen during debug in your Visual Studio.

Answer 4

StreamReader accepts Stream type of object while you are passing it a string. try this,

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");
        }