WinForm开发之“OpenFileDialog”

MSDN路径:.Net开发-》.Net Framework SDK文档-》类库参考-》System.Windows.Forms-》OpenFileDialog类

private void button1_Click(object sender, System.EventArgs e)
{
    Stream myStream;
    OpenFileDialog openFileDialog1 = new OpenFileDialog();

    openFileDialog1.InitialDirectory = "c:\\" ;
    openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ;
    openFileDialog1.FilterIndex = 2 ;
    openFileDialog1.RestoreDirectory = true ;

    if(openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        if((myStream = openFileDialog1.OpenFile())!= null)
        {
            // Insert code to read the stream here.
            myStream.Close();
        }
    }
}

posted @ 2011-05-16 20:30  Pickuper  阅读(1517)  评论(0)    收藏  举报