Geek asks and answers

Does OpenFileDialog InitialDirectory open a relative path?

dialogis a class object OpenFileDialogand I am using a method ShowDialog().

When I use a path containing a relative path, for example:

dialog.InitialDirectory = "..\\abcd";
dialog.InitialDirectory = Directory.GetCurrentDirectory() + "..\\abcd";

ShowDialog()I can only make a specific path starting from disk:

dialog.InitialDirectory = "C:\\ABC\\DEF\\abcd";

In this case, I want the path to be 1 level higher than my current .exe directory, and then down to the directory abcd) The
current .exe path can be found using Directory.GetCurrentDirectory(), which is fine, but I can not continue with "..")

Directory Hierarchy:

ABC
    DEF 
        abcd (where i want)
        defg (where .exe is at)

So, is there a way to use "..\\"with InitialDirectory?
Or should I use a specific path with it?
Thank!

+3
source share
2 answers

!

string CombinedPath = System.IO.Path.Combine(Directory.GetCurrentDirectory(), "..\\abcd");
dialog.InitialDirectory = System.IO.Path.GetFullPath(CombinedPath);
+3

, :

dialog.InitialDirectory
  = Path.Combine(Path.GetDirectoryName(Directory.GetCurrentDirectory()), "abcd");

Path.GetDirectoryName , .

+2

More articles:


All Articles