將MainWindow.xaml中的T​​extBox名稱參數傳遞給button_click函數

Question

我正在製作一個使用Microsoft.Win32.OpenFileDialog的窗體dlg = new Microsoft.Win32.OpenFileDialog（）;提供文件選擇菜單。

我想使用相同的功能來更新輸入文件的文本框和輸出文件的文本框。

<StackPanel Orientation="Horizontal" HorizontalAlignment="Left" Width="515" Height="96" VerticalAlignment="Top"> 
     <TextBlock Text="Input File:" VerticalAlignment="Center" /> 
     <TextBox x:Name="InputFileBox" Width ="213" VerticalAlignment="Center" TextChanged="InputFileBox_TextChanged" Height="17" Margin="0,39,0,40" /> 
     <Button Content="Browse" Width="47" Margin="0,39,0,40" RenderTransformOrigin="1.599,0.714" Click="Browse_Click"/> 
    </StackPanel> 


    <StackPanel Orientation="Horizontal" HorizontalAlignment="Left" Width="515" Height="96" VerticalAlignment="Top"> 
     <TextBlock Text="Output File:" VerticalAlignment="Center" /> 
     <TextBox x:Name="OutputFileBox" Width ="213" VerticalAlignment="Center" TextChanged="OutputFileBox_TextChanged" Height="17" Margin="0,39,0,40" /> 
     <Button Content="Browse2" Width="47" Margin="0,39,0,40" RenderTransformOrigin="1.599,0.714" Click="Browse_Click"/> 
    </StackPanel> 

所以我希望能夠發送「InputFileBox」或「OutputFileBox」與「BrowseClick」，讓我不必有BrowseInputClick和BrowseOutputClick功能。

在Browse_Click功能

我希望能夠做一些事情，如：

private void Browse_Click(object sender, RoutedEventArgs e) 
    { 
     // Create OpenFileDialog 
     Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog(); 


     // Display OpenFileDialog by calling ShowDialog method 
     Nullable<bool> result = dlg.ShowDialog(); 


     // Get the selected file name and display in a TextBox 
     if (result == true) 
     { 
      // Open document 
      string filename = dlg.FileName; 

      // I don't know what to put here: input/outputTextBoxName = filename 
     } 

感謝

Answer 1

在WPF中你可以設置一個按鈕的標籤屬性。

一旦你有，你可以使用

在XAML添加Tag="input"作爲對inputTextBox和Tag="output"到outputTextBox的一個屬性（例如：<TextBox x:Name="inputTextBox" Tag="input"/>）獲得在單擊處理標記屬性

var tag = (sender as Button).Tag; 

然後：

if (tag == 'input') 
    inputTextBox.Text = filename; 
else outputTextBox.Text = filename;