Question

c#.net openfiledialog对话框怎样设置打开文件的默认路径？

Answer 1

private void OpenFile(object sender, EventArgs e)
{
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.Personal);
openFileDialog.Filter = "文本文件(*.xml)|*.xml|所有文件(*.*)|*.*";
if (openFileDialog.ShowDialog(this) == DialogResult.OK)
{
string FileName = openFileDialog.FileName;

private void OpenFile(object sender, EventArgs e)
{
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.Personal);
openFileDialog.Filter = "文本文件(*.xml)|*.xml|所有文件(*.*)|*.*";
if (openFileDialog.ShowDialog(this) == DialogResult.OK)
{
string FileName = openFileDialog.FileName;

//NewDesignView();
//frmDesign.OpenXml(FileName);

}
}
}
}

追问

我需要添加OpenFile方法吗？
我是新手，麻烦在关键地方添加一些注释，谢谢

Answer 2

InitialDirectory属性

追问

这样不行吧，路径名是从配置文件app.config中读取的。

追答

你是做什么用 不是点击按钮
打开一个对话框 这个对话框的默认路径么？

追问

是的。