OpenFileDialog 用 exe 打开文件夹

Question

如何使用 OpenFileDialog 快速打开应用程序的文件夹？

        OpenFileDialog openFileDialog1 = new OpenFileDialog();
        openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
           ...........

        }

Answer 1

猜测您的意思是“从我的应用程序文件夹中开始显示 OpenFileDialog”，只需OpenFileDialog.InitialDir在显示OpenFileDialog.

string AppPath = Path.GetDirectoryName(Application.ExecutablePath);;
openFileDialog1.InitialDir = AppPath;

如果您需要帮助查找应用程序的目录，请参阅获取应用程序的根文件夹