Question

新手提问 C# OpenFileDialog问题 建立一个打开excel的路径

新手提问 C# OpenFileDialog问题 建立一个打开excel的路径
void OpenFileDialog1FileOk(object sender, System.ComponentModel.CancelEventArgs e)
{
OpenFileDialog openFileDialog=new OpenFileDialog();
openFileDialog.InitialDirectory="c:\\";
openFileDialog.Title = "选择导入的文件";
openFileDialog.Filter="excel file|*.*|.xls|所有文件|*.*";
openFileDialog.FilterIndex=1;
if (openFileDialog.ShowDialog()==DialogResult.OK)
{
this.path.Text = openFileDialog.FileName;
}
}

貌似就是 this.path.Text = openFileDialog.FileName; 这句出问题 那里错了 望高手解答

Answer 1

openFileDialog.Filter = "excel file|*.xls|所有文件|*.*";
设置文件扩展名错了
我试过了 应该这样 你试试