ShowDialog with Parent still enabled

Question

In one of our apps we want to want to limit the user from opening other menu items when an existing menu item is already open. We are currently doing this:

    private void menuItem1_Click(object sender, EventArgs e)
    {
         Myform f = new MyForm();
         f.ShowDialog(this);
    }

However in doing this, we lose the ability to interact at all with the parent window because internally, the parent.enabled property was set to false. Using the code above, if the user has menu item open and wants to move the parent window to see something on their desktop, they first must close the menu item, move the parent, and reopen the menu item.

I have come up with the follow method of doing the UI in a backgroundworker

public class BaseForm : Form
{
    private bool _HasChildOpen;
    protected BackgroundWorker bgThead;

    public BaseForm()
    {
        _HasChildOpen = false;

        bgThead = new BackgroundWorker();
        bgThead.DoWork += new DoWorkEventHandler(OpenChildWindow);
        bgThead.RunWorkerCompleted += new System.ComponentModel.RunWorkerCompletedEventHandler(this.ClearChildWindows);
    }


    protected void ClearChildWindows(object sender, RunWorkerCompletedEventArgs e)
    {
        _HasChildOpen = false;
    }

    public void OpenChildWindow(object sender, DoWorkEventArgs e)
    {
        if (!_HasChildOpen)
        {
            Form f = (Form)e.Argument;
            f.StartPosition = FormStartPosition.CenterScreen;
            f.ShowDialog();
        }
    }
}

and then each menu item has the following code

    private void menuItem1_Click(object sender, EventArgs e)
    {
        if (!bgThead.IsBusy)
        {
            bgThead.RunWorkerAsync(new Myform());
        }
    }

but this approach is a big no no. However, using invoke seems to get me back where I started:

    private void doUIWork(MethodInvoker d)
    {
        if (this.InvokeRequired)
        {
            this.Invoke(d);
        }
        else
        {
            d();
        }
    }

    public void OpenChildWindow(object sender, DoWorkEventArgs e)
    {
        if (!_HasChildOpen)
        {

            doUIWork(delegate() {
                Form f = (Form)e.Argument;
                f.StartPosition = FormStartPosition.CenterScreen;
                f.ShowDialog();
            });

            //Form f = (Form)e.Argument;
            //f.StartPosition = FormStartPosition.CenterScreen;
            //f.ShowDialog();
        }
    }

How do I properly limit the user to just one menu item open, but at the same time leave the parent enabled such that it can be moved resized etc?

Dialogs are evil and should always be avoided. But sometimes [you *can't* avoid them](http://stackoverflow.com/a/5183623/17034). Then trying to work around them is just plain foolish. If you must, iterate the parent's Controls collection and set their Enabled property to false. And pinvoke EnableWindow() to re-enable the parent window. — Hans Passant, Dec 03 '13 at 19:14

score 0 · Answer 1 · answered Dec 04 '13 at 21:05

You will need to programmatically disable menu strip behavior once one of the forms is open. So if you have Form1 and Form2, (with a menuStrip on Form1 and toolStripMenuItem1, toolStripMenuItem2 on the menuStrip):

    private void menuItem1_Click(object sender, EventArgs e)
    {
        var f2 = new Form2();
        f2.FormClosing += f2_FormClosing;
        f2.Show();
        this.menuStrip1.Enabled = false;
    }

    private void menuItem2_Click(object sender, EventArgs e)
    {
        var f2 = new Form2();
        f2.FormClosing += f2_FormClosing;
        f2.Show();
        this.menuStrip1.Enabled = false;
    }

    void f2_FormClosing(object sender, FormClosingEventArgs e)
    {
        this.menuStrip1.Enabled = true;
    }

using the Show() method instead of ShowDialog() enables interaction with the parent control, though you will need to manually disable/enable behavior depending on when the child control is shown or not.

ShowDialog with Parent still enabled

1 Answers1