Question

当我单击按钮打开文件时，OpenFileDialog框会打开两次。第一个框仅打开图像文件和第二个框文本文件。如果用户决定在不选择文件的情况下关闭第一个框，则也会弹出第二个框。我不确定我在这个问题上的看法。任何帮助，将不胜感激。谢谢！

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

您正在拨打ShowDialog()两次：

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

只需调用一次，然后保存结果：

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)

编辑：

如果您希望第二个对话框仅在处理第一个对话框时显示，您可以执行以下操作：

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

您正在ShowDialog和if条件中调用else if。该方法负责显示对话框，并且只有很好的副作用，也告诉您用户点击了什么。

将方法的结果存储在变量中，并在if..elseif条件中检查它。

c＃/ WPF openFileDialog框打开两次

2 个答案: