显示一个OpenFileDialog并从一行中获取FileName属性

Question

I'm working on a simple application and I'm trying to get a shortcut to an exe that the user chooses. It's working as it is, but I was wondering if it's possible to open the dialog box and get the file path from it in one line so that I don't have to store the dialog box in memory.

I know it's possible to call multiple methods on one line, such as string[] array = value.Trim().ToLower().Split('\\\\'); but when I try to do that type of setup with the OpenFileDialog, I get an error about the methods not existing.

Here's the code I have right now:

OpenFileDialog box = new OpenFileDialog(); box.ShowDialog(); pathTextBox.Text = d.FileName;

I was wondering if it would be possible (for neatness sake) to set it up something like pathTextBox.Text = new OpenFileDialog().ShowDialog().FileName;

Answer 1

Short answer: it is called method call chaining. It works with Trim().ToLower().Split() because Trim() and ToLower() return string . You cannot chain call to ShowDialog this way, because ShowDialog method returns DialogResult which is just an enum .

However, in theory you could extract this to a separate extension method:

public static class OpenFileDialogExtensions
{
    public static string ShowDialogAndReturnFileName(this OpenFileDialog dialog)
    {
        // consider checking arguments
        dialog.ShowDialog(); 
        // consider checking the result of `ShowDialog` (e.g., user could close the dialog)
        return dialog.FileName;
    }
}

// Usage, just like you want:
pathTextBox.Text = new OpenFileDialog().ShowDialogAndReturnFileName();

Keep in mind that shorter code doesn't mean better code .
Perhaps, the best way to do this is not to do this.

Answer 2

I'm fairly confident that none of the modal dialogs don't work that way. While you can say:

var foo = new OpenFileDialog().ShowDialog();

The result of that is a DialogResult , and not the filename property you're looking for. Additionally, you would no longer have a reference to the actual FileDialog object and could no longer extract the chosen filename anyway.

One alternative available to you is to make a method that makes it "look" like you're doing it with a single call:

public static string GetFilename()
{
    var dlg = new OpenFileDialog();
    var result = dlg.ShowDialog();
    var filename = dlg.FileName;

    return filename;
}

public static void Main()
{
    var userChosenFile = GetFilename();
    var aDifferentChosenFile = GetFilename();
    var yetAnotherChosenFile = GetFilename();
}