Parameter is not valid - "pictureBox.Show()"

I'm making a TIFF viewer with a picture box and buttons for going trought the pages of multi page files.

When I open a file for the first time, all works great, then I close it. But when I open a new file, after closing the previous one, I get an error saying "Parameter is not valid" at the line "pbtiff.Show()".

This is my code for opening the file:

private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
pbtiff.Hide();
OpenFileDialog dialogo = new OpenFileDialog();
dialogo.Title = "Search files";
dialogo.InitialDirectory = @"E:\";
dialogo.Filter = "PDF Files(.pdf)|*.pdf|Image Files (.bmp,.jpg,.png,.tiff,.tif) |*.bmp;*.jpg;*.png;*tiff;*tif";
DialogResult answer = dialogo.ShowDialog();
if (answer == DialogResult.OK)
{
string fullPath = dialogo.FileName;
openFile = fullPath;
using (var fileStream = new FileStream(ficheiroaberto, FileMode.Open))
{
type = Path.GetExtension(caminhoCompleto);
if (type == ".tif" || type == ".tiff")
{
up.Visible = true;
up.Enabled = false;
down.Visible = true;
pbtiff.Show(); -> the error is here, when I open the file for the 2nd time
if (pbtiff.Image != null)
{
pbtiff.Image.Dispose();
}
pbtiff.Image = System.Drawing.Image.FromStream(fileStream);
SplitTiffFinal(fileStream);
filestiff = GetFilesFinal();
up.Visible = true;
up.Enabled = false;
down.Visible = true;
}
}
}
}

Can you help me figure out what is the probem, if it works the first time?

2 Replies

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Sofia Rodrigues7y ago
@Sanwar, I can´t use Dispose() after opening an image because it will open and dispose it afterwords, so the image won't appear at all.
Sanwar Ranwa7y ago
1. pbtiff.Image = System.Drawing.Image.FromStream(fileStream);
2. SplitTiffFinal(fileStream);
3. filestiff = GetFilesFinal();
4. up.Visible = true;
5. up.Enabled = false;
6. down.Visible = true;
7. pbtiff.Image .Dispose();
Try this it may help you