OpenFileDialog 获取文件路径

Question

            if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                textBox2.Text = openFileDialog1.FileName;
            }

Answer 1

一楼都回答完了。放个openFileDialog上去就完事了。
不想放那个对话框上去就加在IF前面加上一句：               OpenFileDialog openFileDialog1 = new OpenFileDialog();
            if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                textBox2.Text = openFileDialog1.FileName;
            }

Answer 2

其实你也可以用OpenFileDialog的FileOk事件实现同样的功能,代码如下:
        private void button1_Click(object sender, EventArgs e)
        {
            OpenFileDialog dialogOpen = new OpenFileDialog();
            dialogOpen.FileOk += delegate(object source, CancelEventArgs args)
            {
                textBox1.Text = dialogOpen.FileName;
            };
            dialogOpen.ShowDialog();
        }

Answer 3

OpenFileDialog ofd = new OpenFileDialog();//new一个方法
ofd.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments);  //定义打开的默认文件夹位置
if (ofd.ShowDialog() == DialogResult.OK) 
{
textBox1.Text = ofd.FileName;               //获得选择的文件路径
extendedName = Path.GetExtension(fileName);       //获得文件扩展名
fileName1 = Path.GetFileName(fileName);           //获得文件名
}

Answer 4

都被他们抢光了，补充一点吧，在button的单击事件里写