WPF, How can I call an OpenFolderDialog and load the content of a folder or subfolders in a listbox ...

Question

Good morning dear friends.
I have a small project. It is a loader of text file names in a ListBox and that in turn allows me to select them and then be able to record or write them in a * .INI file that is already selected with its path.

The "Open Folder" button shows an OpenFolderDialog and loads the path of the selected folder in a TextBox and this in turn loads in the LisBox all the text files.txt located in the chosen folder and also in the subfolders of the main folder .

The texts selected in the ListBox are written or saved in a TextList.ini file (for example), and those that are not selected are deleted from the TextList.ini file. Please I hope you can help me, I have also created a video so that I can have a better reference and I have attached some images as well.

Video Link: https://www.youtube.com/watch?v=bfoQ0GijDgk

Answer 1

You can use both OpenFolderDialog and FolderBrowserDialog after you adding the System.Windows.Forms.dll in the project's reference.
Code to get the folder location:

 private void btnOpen_Click(object sender, RoutedEventArgs e)  
        {  
            System.Windows.Forms.FolderBrowserDialog openFileDlg = new System.Windows.Forms.FolderBrowserDialog();  
            var result = openFileDlg.ShowDialog();  
            if (result.ToString() != string.Empty)  
            {  
                txtPath.Text = openFileDlg.SelectedPath;  
            }  
            root = txtPath.Text;  
        }  

Code to get the txt in folder and subfolders:

private void Button_Click(object sender, RoutedEventArgs e)  
        {  
            lt = new ObservableCollection<MyModel>();  
            string[] dicFileList = Directory.GetFiles(root, "*.txt", SearchOption.AllDirectories);  
            foreach (string element in dicFileList)  
            {  
                myModel = new MyModel();  
                //myModel.Name = System.IO.Path.GetFileName(element);  
                myModel.Name = System.IO.Path.GetFileNameWithoutExtension(element);  
                myModel.StatusForCheckBox = false;  
  
                lt.Add(myModel);  
            }  
            myList.ItemsSource = lt;  
        }  

You can create a field public bool StatusForCheckBox { get; set;} for MyModel to identity selected or not:

 private void Save_Click(object sender, RoutedEventArgs e)  
        {  
            List<string> ltForSave = new List<string>();  
            foreach (MyModel obj in myList.ItemsSource)  
            {        
                if(obj.StatusForCheckBox==false)  
                {  
                    ltForSave.Add(";" + obj.Name);  
                }else  
                {  
                    ltForSave.Add(obj.Name);  
                }  
            }  
            //Save ltForSave as ini file.      
        }  

Use the listbox like:

<ListBox x:Name="myList" Width="400" Height="400" HorizontalAlignment="Right">  
            <ListBox.ItemTemplate>  
                <DataTemplate >  
                    <CheckBox Content="{Binding Name}" IsChecked="{Binding StatusForCheckBox,Mode=TwoWay}"/>  
                </DataTemplate>  
            </ListBox.ItemTemplate>  
        </ListBox>  

---

If the response is helpful, please click "Accept Answer" and upvote it.
Note: Please follow the steps in our documentation to enable e-mail notifications if you want to receive the related email notification for this thread.

Answer 2

This doesn't work for me in Visual Studio 2022. I found a reference to System.Windows.Forms.VisualStudio.15.0.dll, so I added that. Trouble is, it didn't work. When I try to put in a using statement, it doesn't recognize System.Windows.Forms.