R
rk
According to the docs, I would expect to check the result of an
OpenFileDialog in a VS 2005 CLR Windows Forms application this way:
if (openFileDialog1->ShowDialog()==
DialogResult::OK)
However, this does not work, and I have to use:
if (openFileDialog1->ShowDialog()==
System::Windows::Forms:ialogResult::OK)
Did I miss anything, or is this a bug?
Thanks
Richard
OpenFileDialog in a VS 2005 CLR Windows Forms application this way:
if (openFileDialog1->ShowDialog()==
DialogResult::OK)
However, this does not work, and I have to use:
if (openFileDialog1->ShowDialog()==
System::Windows::Forms:ialogResult::OK)
Did I miss anything, or is this a bug?
Thanks
Richard