private void OpenThreeFile()
{
this.openFileDialog1.FileName = System.IO.Path.GetDirectoryName(this.lstFiles.SelectedItem.ToString());
string openFileName;
openFileName = this.openFileDialog1.FileName;
try
{
System.Diagnostics.Process.Start(openFileName);
this.btnOpenThreeFile.Enabled = false;
}
catch (Exception ex)
{
MessageBox.Show(ex.Message, "错误提示", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
}
问题:为什么openFileName 获取不了lstFiles的值?
哪里错了啊?
![图片说明](https://img-ask.csdn.net/upload/201710/10/1507647142_924343.jpg)