Allowed Menustrip only one child at moment

How can i allow only one window at a time when i click menu in menustrip ?

Ex: i have Menustrip Ordre, Tarif, etc... when i click Ordre for the first time it will open a new form, but for the second time i want to disallow it.

 private void ordresToolStripMenuItem_Click(object sender, EventArgs e)
    {

        if (Already open)
        {

        }
        el开发者_运维百科se
        {
            Lordre newMDIChild = new Lordre(ClientId);
            // Set the Parent Form of the Child window.
            newMDIChild.MdiParent = this;
            // Display the new form.
            newMDIChild.Show();                
        }

    }

Thanks you in advance

---

If you want the form to be created only the first time, and then show that same form the next time the menu item is selected, something like this could work:

private Lordre orderForm = null;
private void ordresToolStripMenuItem_Click(object sender, EventArgs e)
{
    if (orderForm == null)
        orderForm = new Lordre(ClientId);
        // Set the Parent Form of the Child window.
        orderForm .MdiParent = this;

    }
    // Display the form.
    orderForm.Show(); 
    orderForm.Activate();
}

---

Maybe something like this:

public class MyForm
{
    private Window _openedWindow;
    private void ordresToolStripMenuItem_Click(object sender, EventArgs e)
    {

        if ( _openedWindow != null &&  _openedWindow.Open)
        {
            //
        }
        else
        {
            Lordre newMDIChild = new Lordre(ClientId);
            _openedWindow = newMDIChild;
            // Set the Parent Form of the Child window.
            newMDIChild.MdiParent = this;
            // Display the new form.
            newMDIChild.Show();                
        }
    }
}

This was written entirely in the browser and I haven't written a windows app for a long time, so the exact classes and properties may be different.

---

The way I usually handle single instance forms is just to have a member variable to hold it, and then check to see whether it is null.

so, have a member variable:

private TestForm myTestForm = null;

and then when you are checking just check whether it's null; if not, when you create your form assign it to your member variable and attach to the event handler for the closing event of the child form.

if (myTestForm != null)
{
   MessageBox.ShowDialog("Sorry, you already have a TestForm open!");
}
else
{
   myTestForm = new TestForm();
   myTestForm.FormClosing += myTestForm_FormClosing;
   myTestForm.MdiParent = this;
   myTestForm.Show();
}

and in the closing handler, just set it back null.

private void myTestForm_FormClosing(Object sender, FormClosingEventArgs e)
{
   myTestForm = null;
}

also, I did a bit of searching and rather than having the FormClosing event and handler, you can just change your conditional to be:

if ((myTestForm != null) && (!myTestForm.IsDisposed())

---

Thanks you all for your response.

i have found this one

private void ordresToolStripMenuItem_Click(object sender, EventArgs e)
    {
        // a flag to store if the child form is opened or not
        bool found = false;   
        // get all of the MDI children in an array
        Form[] charr = this.MdiChildren;

        if (charr.Length == 0)     // no child form is opened
        {
            Lordre myPatients = new Lordre();
            myPatients.MdiParent = this;
            // The StartPosition property is essential
            // for the location property to work
            myPatients.StartPosition = FormStartPosition.Manual;
            myPatients.Location = new Point(0,0);
            myPatients.Show();
        }
        else     // child forms are opened
        {
            foreach (Form chform in charr)
            {
                if (chform.Tag.ToString() == "Ordre")  // one instance of the form is already opened
                {
                    chform.Activate();
                    found = true;
                    break;   // exit loop
                }
                else
                    found = false;      // make sure flag is set to false if the form is not found
            }

            if (found == false)    
            {
                Lordre myPatients = new Lordre();
                myPatients.MdiParent = this;
                // The StartPosition property is essential
                // for the location property to work
                myPatients.StartPosition = FormStartPosition.Manual;
                myPatients.Location = new Point(0,0);
                myPatients.Show();
            }
        }  
    }

it is what i want, but it is too much code. i wonder if i can reduce it.

i have to make this to each of my stripmenu.

if (chform.Tag.ToString() == "Ordre")

if (chform.Tag.ToString() == "Another one")

---

The way I do it is by first storing the screen in a variable in a class like so:

private Lordre _LordreChildForm = new Lordre(ClientID)

Then make a property for it like so:

public Lordre LordreChildForm { get => _LordreChildForm; set => _LordreChildForm = 
value; }

Then on the form's Form Closing event set the value to null like so:

private void Lordre_FormClosing(object sender, FormClosingEventArgs e)
{
            ClassName.LordreChildForm= null;
}

Now Finally you can set an if statement in the button click event handler like so:

private void LordreToolStripMenuItem_Click(object sender, EventArgs e)

{
    if(ClassName.LordreChildForm== null || ClassName.LordreChildForm.IsDisposed)
    {
        ClassName.LordreChildForm= new DeregisterClub();
        ClassName.LordreChildForm.MdiParent = this;
    }
    LordreChildForm.Dock = DockStyle.Fill;
    LordreChildForm.Show();
    LordreChildForm.Focus();
}