ShowDialog() 错误的解决

showdialog() 错误的解决

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