OpenFileDialog 类的ShowDialog() 错误的解决

首先，一个类里，有个linkLabel1

private OpenFileDialog openFileDialog1;
private DialogResult result;

private void linkLabel1_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
        {

            openFileDialog1 = new OpenFileDialog();
            string patch = Application.StartupPath + "\\LOG\\";
            openFileDialog1.InitialDirectory = patch;
            openFileDialog1.Filter = "xls files (*.xls)|*.xls";

            result = openFileDialog1.ShowDialog();

            if (result == DialogResult.OK)
            {
                if (openFileDialog1.FileName != "")
                {
                    Process.Start(openFileDialog1.FileName);
                }
               
            }

           
        }

就会报 
在可以调用 OLE 之前，必须将当前线程设置为单线程单元(STA)模式。请确保您的 Main 函数带有 STAThreadAttribute 标记。 只有将调试器附加到该进程才会引发此异常。

在测试小程序里没有问题，当移到大程序里就这样的问题了。可能是线程多的原因。解决办法就是添加线程，代码如下

private Thread invokeThread;

private void linkLabel1_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
        {
            openFileDialog1 = new OpenFileDialog();
            openFileDialog1.InitialDirectory = patch;
            openFileDialog1.Filter = "xls files (*.xls)|*.xls";

            invokeThread = new Thread(new ThreadStart(InvokeMethod));
            invokeThread.SetApartmentState(ApartmentState.STA);
            invokeThread.Start();
            invokeThread.Join();

            if (result == DialogResult.OK)
            {
                if (openFileDialog1.FileName != "")
                {
                    Process.Start(openFileDialog1.FileName);
                }

            }
        }

private void InvokeMethod()
        {
            result = openFileDialog1.ShowDialog();
        }

问题得到解决