Let's say I have a file path of:
C:\Users\my_name\Desktop\my_project\bin\debug\my_project.exe
How can I get the following file path to use in an OpenFileDialog
?
C:\Users\my_name\Desktop\my_project\
EDIT:
I've tried the following..
var path = Directory.GetParent(Directory.GetCurrentDirectory()).Parent?.ToString();
var path = Assembly.GetExecutingAssembly().CodeBase;
But both return incorrect values.
You can use the following logic:
DirectoryInfo Di = Directory.GetParent(@"C:\Users\my_name\Desktop\my_project\bin\debug\my_project.exe");
// Which will give you the debug folder
int DirectoryLevel = 3;
for (int i = 1; i < DirectoryLevel; i++)
{
Di = Di.Parent;
// Which will give you the bin fodler when i =1
// Which will give you the my_project folder when i =2
}
string currentDirectory = Di.FullName; // Give the path