c＃/ WPF openFile对话框打开两次

Question

When I click the button to open a file the OpenFileDialog box opens twice. The first box only opens the image files and the second box text files. If the user decides to close out the first box without choosing a file the second box pops up as well. I'm not sure what I am over looking on this issue. Any help would be appreciated. Thanks!

OpenFileDialog of = new OpenFileDialog();   
of.Filter = "All Image Formats|*.jpg;*.png;*.bmp;*.gif;*.ico;*.txt|JPG Image|*.jpg|BMP image|*.bmp|PNG image|*.png|GIF Image|*.gif|Icon|*.ico|Text File|*.txt";

if (of.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
try
{
   image1.Source = new BitmapImage(new Uri(of.FileName));

    // enable the buttons to function (previous page, next page, rotate left/right, zoom in/out)
    button1.IsEnabled = false;
    button2.IsEnabled = false;
    button3.IsEnabled = false;
    button4.IsEnabled = false;
    button5.IsEnabled = true;
    button6.IsEnabled = true;
    button7.IsEnabled = true;
    button8.IsEnabled = true;                          
}
catch (ArgumentException)
{
    // Show messagebox when argument exception arises, when user tries to open corrupted file
    System.Windows.Forms.MessageBox.Show("Invalid File");
}
 }
 else if(of.ShowDialog() == System.Windows.Forms.DialogResult.OK)
 {
try
{
    using (StreamReader sr = new StreamReader(of.FileName))
    {
        textBox2.Text = sr.ReadToEnd();
        button1.IsEnabled = true;
        button2.IsEnabled = true;
        button3.IsEnabled = true;
        button4.IsEnabled = true;
        button5.IsEnabled = true;
        button6.IsEnabled = true;
        button7.IsEnabled = true;
        button8.IsEnabled = true;

    }
}
catch (ArgumentException)
{
    System.Windows.Forms.MessageBox.Show("Invalid File");
}
} 

Answer 1

You're calling ShowDialog() twice:

if (of.ShowDialog() == System.Windows.Forms.DialogResult.OK)
    {
       //...
    }
    else if(of.ShowDialog() == System.Windows.Forms.DialogResult.OK)

Just call it once, and save the results:

var dialogResult = of.ShowDialog();
if (dialogResult == System.Windows.Forms.DialogResult.OK)
{
       //...
}
// Note that the condition was the same!
else if(dialogResult != System.Windows.Forms.DialogResult.OK)

---

Edit:

If you want the second dialog to only show when the first is handled, you can do:

var dialogResult = of.ShowDialog();
if (dialogResult == System.Windows.Forms.DialogResult.OK)
{
       //...

    // Do second case here -  Setup new options
    dialogResult = of.ShowDialog(); //Handle text file here
}

Answer 2

You are calling ShowDialog both in the if and else if condition. That method is responsible for showing the dialog and only has the nice side effect of also telling you what the user clicked.

Store the method's result in a variable and check that in your if..elseif conditions instead.