在Console Application中浏览文件夹

Question

I currently have to code to allow me to read all of the files of a folder and write them to the console. Below, I also have got the code to select individual files from a directory using a browser. I would like to know how I would be able to select a folder using a browse button.

code to check all files

  foreach(var path in Directory.GetFiles(@"C:\Name\Folder\"))
    {
       Console.WriteLine(path); // full path
       Console.WriteLine(System.IO.Path.GetFileName(path)); // file name
    }

Code to open dialog box

OpenFileDialog fileSelectPopUp = new OpenFileDialog();
            fileSelectPopUp.Title = "";
            fileSelectPopUp.InitialDirectory = @"c:\";
            fileSelectPopUp.Filter = "All EXCEL FILES (*.xlsx*)|*.xlsx*|All files (*.*)|*.*";
            fileSelectPopUp.FilterIndex = 2;
            fileSelectPopUp.RestoreDirectory = true;
            if (fileSelectPopUp.ShowDialog() == DialogResult.OK)
            {
                textBox1.Text = fileSelectPopUp.FileName;
            }

Answer 1

First you need to add reference to System.Windows.Forms

Then, Add STAThread Attribute to the main method. This indicates that your program is single-threaded and enabled it to work with COM components (which the System dialogs use).

After that only you can use the FolderBrowserDialog with the Console Application

static class Program
{
    [STAThread]
    static void Main(string[] args)
    {
        FolderBrowserDialog fbd = new FolderBrowserDialog();
        if (fbd.ShowDialog() == DialogResult.OK)
        {
            foreach (var path in Directory.GetFiles(fbd.SelectedPath))
            {
                Console.WriteLine(path); // full path
                Console.WriteLine(System.IO.Path.GetFileName(path)); // file name
            }
        }


    }
}

Answer 2

User the FolderBrowserDialog

FolderBrowserDialog b = new FolderBrowserDialog();

if(b.ShowDialog() == DialogResult.OK)
{
  var folderName = b.SelectedPath;
}

Answer 3

虽然，为图像UI操作，你可以使用DotImaging.UI ：

string fileName = UI.OpenFile(); //open-file dialog