The calling thread must be STA, because many UI components require this error In WPF?

Question

Hi.

I make pop-up message for my application. this form is on wpf, and in the a thread i check a value and for some result , I show that pop-up to show message.

In the thread when I create pop-up class(type of WPF window class) an error throw by pop-up Contractor. "The calling thread must be STA, because many UI components require this"

What do I do?

//-------new edit :(

I use this for show my form in thread but still have error !!(cuz thread.start() in the timer thread)

System.Threading.Thread Messagethread = new System.Threading.Thread(
                new System.Threading.ThreadStart(delegate()
                    {
                        System.Windows.Threading.DispatcherOperation DispacherOP = frmMassenger.Dispatcher
                            .BeginInvoke(System.Windows.Threading.DispatcherPriority.Normal, new Action(
                                delegate()
                                {
                                       frmMassenger.Show();
                                }
                        ));
                    } ));
            Messagethread.Start();

Answer 1

For your thread that you're trying to start the GUI element in, you need to set the appartment state of your thread to STA BEFORE you start it.

Example:

myThread.SetApartmentState(ApartmentState.STA);
myThread.Start();

Answer 2

Or put the STAThread attribute on your program's entry point method (usually Main):

[STAThread]
public static void Main(string[] args)
{
    ...
}

Answer 3

Absolutely Dispatcher is only way to do something(in specific Thread) when we work with multi-threading in wpf!
but for work with Dispatcher we must know 2 things:
1)too many way to use Dispatcher like Dispatcher_Operation , [window.dispatcher] ,...
2)We must call dispatcher in the main thread of app(that thread is STA thread)

so for example if u want show other window[wpf] in another thread u can use this code:

Frmexample frmexample = new Frmexample();
            Frmexample .Dispatcher.BeginInvoke
                (System.Windows.Threading.DispatcherPriority.Normal, (Action)(() =>
                {
                    frmexample.Show();//-----or do any thing u want with that form
                }
                ));

tip: remember u cant access any fields or properties from out dispatcher, so use that wisely