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Home/ Questions/Q 8099775
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T22:31:18+00:00 2026-06-05T22:31:18+00:00

I have a simple issue, but the solution appears to be tricky. I want

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I have a simple issue, but the solution appears to be tricky. I want to print using the WPF control canvas during a loop; but for each iteration, I want to udpate the canvas control.

If I want to print a canvas control in WPF, I can simply call

PrintDialog dialog = new PrintDialog();
dialog.PrintVisual(this.canvas, "");

And it prints as expected to my default printer. Wonderful.

However, if I want to perform this multiple times in a loop and make an update to the canvas during each iteration, only the final iteration of the loop is printed.

    private void methodName()
    {
        for (int i = 0; i < 2; i++)
        {
            updateTextBox(i.ToString());
            PrintDialog dialog = new PrintDialog();
            dialog.PrintVisual(this.canvas, "");
        }
    }

    private void updateTextBox(string text)
    {
        txtTextBox.Text = text;
    }

Any idea what I need to do to ensure that I get 2 print outs, the first with the txtTextBox.Text value of 0, the second time it has the value of 1?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T22:31:20+00:00Added an answer on June 5, 2026 at 10:31 pm

    OK

    I solved it.

    I removed all the dispatcher object methods so it runs on a single thread.

    To update the canvas, I used the canvas.UpdateLayout() method.

    I also ensured that the print had finished before updating the next canvas (the next iteration).

    private void methodName()
    {
        for (int i = 0; i < 2; i++)
        {
            updateTextBox(i.ToString());
    
            this.canvas.UpdateLayout();
    
            PrintDialog dialog = new PrintDialog();
            dialog.PrintVisual(this.canvas, "ABC");
    
            dialog.PrintQueue.Refresh();
    
            while (dialog.PrintQueue.NumberOfJobs != 0)
            {
                bool isQueued = false;
                foreach (var job in dialog.PrintQueue.GetPrintJobInfoCollection())
                {
                    if (job.Name == "ABC")
                        isQueued = true;
                }
    
                if (!isQueued)
                    break;
    
                Thread.Sleep(500);
                dialog.PrintQueue.Refresh();
            }
        }
    }
    
    private void updateTextBox(string text)
    {
        txtTextBox.Text = text;
    }
    

    I also could have just done thread.sleep(3000) – this worked as it was enough time to ensure the print job had completed, but it was also a little bit ‘hopeful’ and I wanted something more secure.

    Thank you to everyone for your suggestions.

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