[C#]讀檔，彈出DLG，指定檔名

//彈跳出DLG 選擇TXT檔案 讀取字串

ArrayList alWorkPoint = new ArrayList();

//System.Windows.Forms.ListView listView_CM; 宣告的Listview元件

OpenFileDialog dialog = new OpenFileDialog();

dialog.Title = "Select file";
dialog.InitialDirectory = ".\\";
dialog.Filter = "txt files (*.*)|*.txt";
            if (dialog.ShowDialog() == DialogResult.OK)
            {
                StreamReader filer = new StreamReader((dialog.FileName));
                alWorkPoint.Clear();
                listView_CM.Items.Clear();
                string str;
                while ((str = filer.ReadLine()) != null)
                {
                    alWorkPoint.Add(str);
                }
                DecoderFileToListView();
                filer.Close();
            }

//指定檔名，讀取單一檔案

StreamReader filer = new StreamReader(("new.txt"));
string str;
while((str = filer.ReadLine()) != null)
{
    MessageBox.Show(str);
}


測試環境:vs2015 c#.net