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name of form is mp_predracun

If opened show mp_predracun, else open new mp_predracun

What I have tried:

private void PredračunToolStripMenuItem_Click(object sender, EventArgs e)
        {
           
            mp_predracun newMDIChild = new mp_predracun();
            newMDIChild.MdiParent = this;
            newMDIChild.Show();
            
        }
Posted
Updated 12-Apr-19 2:38am
Comments
[no name] 11-Apr-19 13:28pm    
elaborate on your question

1 solution

How about something like ...
C#
static mp_predracun FormToOpen = null;

private void PredračunToolStripMenuItem_Click(object sender, EventArgs e)
        {
                if (FormToOpen == null) // First time in
                        {
                                FormToOpen = new mp_predracun();
                                FormToOpen.MdiParent = this;
                        }

                FormToOpen.Show(); // Show the saved / newly created form
        }

or, more obtuse but shorter ...
C#
static mp_predracun FormToOpen = null;

private void PredračunToolStripMenuItem_Click(object sender, EventArgs e)
        {
                FormToOpen = FormToOpen == null ? new mp_predracun() : FormToOpen;
                FormToOpen.MdiParent = this;
                FormToOpen.Show(); // Show the saved / newly created form
        }
 
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