C＃OpenFileDialog无法打开文件

Question

I'm trying to use openFileDialog to open a Bitmap image and place it on my form. My form construtor...

 public Form1()
    {
        InitializeComponent();
        drawing = new Bitmap(drawingPanel.Width, drawingPanel.Height, drawingPanel.CreateGraphics());
        Graphics.FromImage(drawing).Clear(Color.White);

        // set default value for line thickness
        tbThickness.Text = "5";
    }

... opens a new form with a blank screen, and I can draw on it using the mouse and various color selector buttons. I then save the file with this method:

private void btnSave_Click(object sender, EventArgs e)
    {
        // save drawing
        if (file == null)   // file is a FileInfo object that I want to use
                            // to check to see if the file already exists 
                            // I haven't worked that out yet
        {
            drawing.Save("test.bmp");
            //SaveBitmap saveForm = new SaveBitmap();
            //saveForm.Show();
        }
        else
        {
            drawing.Save(fi.FullName);
        }
    }

The image does save to the debug folder as a .bmp file. Then I use OpenFileDialog to open the file:

private void btnOpen_Click(object sender, EventArgs e)
    {
        FileStream myStream;
        OpenFileDialog openFile = new OpenFileDialog();
        openFile.Filter = "bmp files (*.bmp)|*.bmp";

        if (openFile.ShowDialog() == DialogResult.OK)
        {
            try
            {
                if ((myStream = (FileStream)openFile.OpenFile()) != null)
                {
                    using (myStream)
                    {
                        PictureBox picBox = new PictureBox();
                        picBox.Location = drawingPanel.Location;
                        picBox.Size = drawingPanel.Size;
                        picBox.Image = new Bitmap(openFile.FileName);
                        this.Controls.Add(picBox);
                    }
                }
            }
            catch (Exception ex)
            {

            }
        }
    }

What happes is that OpenFileDialog box comes up. When I select the file test.bmp, the screen goes away and then reappears, when I select it again, the OpenFileDialog window goes away and I'm back to my form with no image. Was hoping for some pointers. No compile or runtime errors.

Answer 1

Why are you calling ShowDialog() twice?

Just call ShowDialog once, so it doesn't open twice , like you indicated.

From MSDN :

OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "bmp files (*.bmp)|*.bmp";

if(openFileDialog1.ShowDialog() == DialogResult.OK)
{
    try
    {
        if ((myStream = openFileDialog1.OpenFile()) != null)
        {
            using (myStream)
            {
                // Insert code to read the stream here.
                PictureBox picBox = new PictureBox();
                picBox.Location = drawingPanel.Location;
                picBox.Size = drawingPanel.Size;
                picBox.Image = new Bitmap (myStream);
                this.Controls.Add(picBox);
            }
        }
    }
    catch (Exception ex)
    {
        MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
    }
}

Answer 2

You open a dialog panel, then when it closes you check to see if the result was OK; then you open another new dialog in the using block; then you assign the image result of that to the PictureBox , and then throw everything away when the using block disposes.

Answer 3

You're calling ShowDialogue twice which is likely the source of your problem. Just use the following code, remove everything else from the method. Your use of using is also incorrect. it does clean up which is disposing of the results. You need to refactor or remove the using statement.

private void btnOpen_Click(object sender, EventArgs e)
{
     OpenFileDialog dlg = new OpenFileDialog()
     {
            dlg.Title = "Open Image";
            dlg.Filter = "bmp files (*.bmp)|*.bmp";

            if (dlg.ShowDialog() == DialogResult.OK)
            {
                PictureBox picBox = new PictureBox();
                picBox.Location = drawingPanel.Location;
                picBox.Size = drawingPanel.Size;
                picBox.Image = new Bitmap (dlg.FileName);
                this.Controls.Add(picBox);
            }
      }
  }

The code above works but is without clean up or error handling. I'll leave that to you.