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Single-threaded apartment - cannot instantiate ActiveX control

I need to get information about applied CSS styles in HTML page. I used AxWebBrowser and iterate IHTMLDOMNode. I'm able to get all the data I need and move the code into my application. The problem is that this part is running inside of the background worker and I got exception when trying to instantiate the control.

AxWebBrowser browser = new AxWebBrowser();

ActiveX control '8856f961-340a-11d0-a96b-00c04fd705a2' cannot be instantiated
because the current thread is not in a single-threaded apartment.

Is there any way how to solve this or other option than AxWebBrowser?

like image 321
martin.malek Avatar asked Sep 13 '09 18:09

martin.malek


3 Answers

The problem you're running into is that most background thread / worker APIs will create the thread in a Multithreaded Apartment state. The error message indicates that the control requires the thread be a Single Threaded Apartment.

You can work around this by creating a thread yourself and specifying the STA apartment state on the thread.

var t = new Thread(MyThreadStartMethod);
t.SetApartmentState(ApartmentState.STA);
t.Start();
like image 98
JaredPar Avatar answered Nov 11 '22 03:11

JaredPar


Go ahead and add [STAThread] to the main entry of your application, this indicates the COM threading model is single-threaded apartment (STA)

example:

static class Program
{
    /// <summary>
    /// The main entry point for the application.
    /// </summary>
    [STAThread]
    static void Main()
    {
        Application.EnableVisualStyles();
        Application.SetCompatibleTextRenderingDefault(false);
        Application.Run(new WebBrowser());
    }
}
like image 43
user764177 Avatar answered Nov 11 '22 01:11

user764177


If you used [STAThread] to the main entry of your application and still get the error you may need to make a Thread-Safe call to the control... something like below. In my case with the same problem the following solution worked!

Private void YourFunc(..)
{
    if (this.InvokeRequired)
    {
        Invoke(new MethodInvoker(delegate()
        {
           // Call your method YourFunc(..);
        }));
    }
    else
    {
        ///
    }
like image 4
Ahmad Avatar answered Nov 11 '22 01:11

Ahmad