Preventing opening duplicate winforms.

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.NET3.0

Hello Experts,
I am facing opening of duplication forms.
here is my code.

C#

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Reflection;

namespace ServiceManagementSystem
{
    public partial class MainCont : Form
    {
        private System.Collections.ArrayList arrMainMenu = new System.Collections.ArrayList();
        private System.Collections.ArrayList arrSubMenu = new System.Collections.ArrayList();
        public System.Windows.Forms.MenuStrip AppMenu;
        public MainCont()
        {
            InitializeComponent();
        }

        private void MainCont_Load(object sender, EventArgs e)
        {
            try
            {
                this.AppMenu = new System.Windows.Forms.MenuStrip();
                // 
                // AppMenu
                // 
                this.AppMenu.BackColor = System.Drawing.SystemColors.GradientActiveCaption;
                this.AppMenu.Font = new System.Drawing.Font("Segoe UI", 11F, System.Drawing.FontStyle.Bold, System.Drawing.GraphicsUnit.Point, ((byte)(0)));
                this.AppMenu.Location = new System.Drawing.Point(0, 0);
                this.AppMenu.Name = "AppMenu";
                this.AppMenu.Size = new System.Drawing.Size(917, 24);
                this.AppMenu.Text = "AppMenu";
                this.Controls.Add(this.AppMenu);
                LoadMenu();
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }
        //Menu loading
        private void LoadMenu()
        {
            arrMainMenu.Add("File");
            arrMainMenu.Add("QCS");
            arrMainMenu.Add("Help");
            arrSubMenu.Add("PSFCalling");
            ToolStripMenuItem tsm = null;
            for (int i = 0; i < arrMainMenu.Count; i++)
            {
                tsm = new ToolStripMenuItem(arrMainMenu[i].ToString());
                int j = 0;
                if (i == 0)
                    j = 0;
                else
                    j = 2;
                for (; j < arrSubMenu.Count; j++)
                {
                    if ((i == 1) && (j == 2))
                        break;
                    tsm.DropDown.Items.Add(arrSubMenu[j].ToString());
                }
                tsm.DropDown.ItemClicked += new ToolStripItemClickedEventHandler(DropDown_ItemClicked);
                this.AppMenu.Items.Add(tsm);
            }
        }

        void DropDown_ItemClicked(object sender, ToolStripItemClickedEventArgs e)
        {
            try
            {
                CallWindow(e.ClickedItem.Text);
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }
        //
        private void CallWindow(string myForm )
        {
            Type type = Type.GetType("ServiceManagementSystem." + myForm);
            Form c = Activator.CreateInstance(type) as Form;
            c.Show();
        }
    }
}

In this Menu Items and sub menu Items are temporally added in to arryList. and all menu generation code is temp. later I will call from database.. I want make admin authority to select menu Items particular User.
I took care for Forms name and Submenu Items text will same.
here I am not understanding how to prevent duplicate forms opening.even though form is in solution I can't use its instance name.waiting for some good solution.

Thanks

Posted 15-Aug-12 3:56am

Basics Learner

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Comments

[no name] 15-Aug-12 10:03am

Do you mean you want to c.ShowDialog() instead?

Basics Learner 15-Aug-12 10:34am

No! it will prevent to opening another window.
need c.Show() but need to prevent duplicate windows.

2 solutions

Solution 2

I agree with BobJ - you need to know what's already open.

Not sure if something like this could work for you....

C#

private bool IsFormOpen(Form frm)
{
    bool retval = false;
    FormCollection fc = Application.OpenForms;
    foreach (Form Appforms in fc)
    {
       if (Appforms == frm)
       {
          retval = true;
       }
    }
    return (retval);
}

Pass the form you think you might want to open into this function....if its already open then you can decide not to open it again.

Aero

Posted 15-Aug-12 5:40am

Aero72

Comments

Basics Learner 15-Aug-12 11:54am

As per logic it should work.I already tried this way. but it always returning false.

Basics Learner 15-Aug-12 11:58am

if I use this way with static form name it is working. in previous application I worked in same way.

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BobJanova · Accepted Answer · 2012-08-15T04:03:00

Solution 1

To prevent opening of a duplicate, you must first know what is open. I'm not sure about the reflective lookup but let's leave that for now, and register windows when you open them:

private void CallWindow(string myForm )
{
    Type type = Type.GetType("ServiceManagementSystem." + myForm);
    Form c;
    if(!openForms.TryGetValue(type, out c) || c.IsDisposed){
      openForms[type] = c = Activator.CreateInstance(type) as Form;
    }
    c.Show();
}

private Dictionary<Type, Form> openForms = new Dictionary<Type, Form>();

Posted 15-Aug-12 4:03am

BobJanova

Updated 15-Aug-12 5:07am

v2

Comments

Basics Learner 15-Aug-12 10:17am

Hi Thanks for quick response.
seems it works. I am getting errors
Error 1 The best overloaded method match for 'System.Collections.Generic.Dictionary<system.type,system.windows.forms.form>.TryGetValue(System.Type, out System.Windows.Forms.Form)' has some invalid arguments
Error 2 Argument '2' must be passed with the 'out' keyword
Error 3 The event 'System.ComponentModel.Component.Disposed' can only appear on the left hand side of += or -=

seems syntax errors

BobJanova 15-Aug-12 10:42am

That's what I get for typing it straight in here ... though you should be able to work it out as well. Try

if(!openForms.TryGetValue(type, out c) || c.IsDisposed)

Basics Learner 15-Aug-12 10:48am

Yes I done same thing.
I was some late here to post. Out of desk :)

Thanks for your help

BobJanova 15-Aug-12 11:07am

Welcome. I edited the solution with these modifications.