Develop Reference

Taskbar icon opens the wrong Active form

Having a strange behaviour with my application whereby when navigation between web browser (or any other application) and clicking back into the application seems to open the wrong Form? So user literally has to make use of Tab window to open the correct Form.
For e.g. Form1 is the main form. User clicks a button which opens Form2. Form1 is hidden behind the scene while Form2 opens. Now if user goes to a different application e.g. browser and clicks back into the application Form1 is displayed and there is no other way of going back to Form2 without the Tab window?
I have used the .ShowDialog() property when opening Form2 which disabled the parent form Form1 but still seems to do it occasionally?!?
I have also set the ShowInTaskBar for Form2 to False so that there is one icon in taskbar for all forms.
Not really sure what can cause this behaviour to happen?
private void button1_Click(object sender, EventArgs e)
{
Form2 form2 = new Form2();
form2.ShowDialog();
}

I think you need to tell Form2 who its owner form is.
Like this:
private void button1_Click(object sender, EventArgs e)
{
Form2 form2 = new Form2();
form2.Owner = this;
form2.ShowDialog();
}
see System.Windows.Forms for more information

When you open form2, you need set mdiParent of form1.
First, set form1 as "isMdiParent" to "true" in properties, and then use the follow code:
private void button1_Click(object sender, EventArgs e)
{
Form2 form2 = new Form2();
form2.MdiParent = this;
form2.Show();
}

Call a third Form from the first Form

I have a "beginner" simple situation here:
i have my main form with a button calling a second form (form2) whit mainform parameters as:
In Form1 :
button_click
Form2 F2 = new Form2(this);
F2.Show();
In Form2 :
public class Form2(Form1 form1)
InitializeComponent(); mainForm = form1;
Ok now i have a Form3(Form1 form1) and i want to call it (show) from Form2 but when i put the code in the second form (Form2):
button_click
Form3 F3 = new Form3(this);
F3.Show();
gives me an error . I tried putting (Form1 form1) instead of (this) but it doesnt work.
How to call Form3 form Form2?

Your attemps show a lack of understanding how parameters are passed to methods, it's not strictly related to winforms.
Anyway, you have declared a Form3 that takes an instance of Form1 as a parameter. If inside your Form2 code you do new Form3(this) the this will reference the instance of the object you're currently in, which is an instance of Form2, and it does not match the form signature.
Also, you can't pass a parameter to a method declaring it's type like you did - new Form3(Form1 form1) - as it does not make any sense and it is not valid syntax.
Since you have stored the Form1 instance reference in a local variable mainForm and your Form3 requires an instance of Form1 you should instantiate it like this: new Form3(mainForm). Make sure the mainForm variable is accessible from where your instantiate Form3.

Form3 F3 = new Form3(mainForm);
F3.Show();

C# WinForms application showing many tray icons

I'm super new to C# and this is my third question here regarding it.
I'm making an app which can be minimized to the system tray. I have two forms named Form1 and Form2. What I have done so far is:
In Form1, I have a button which is showing the Form2 using this code:
this.Hide();
Form2 form2 = new Form2();
form2.Show();
The Form2 has a button which is hiding it, using this code:
this.Hide();
Now, I have the tray icon on the task bar. The tray icon has a ContexMenuStrip, and there is an option named show using this code:
Form1 form1 = new Form1();
form1.Show();
The problem is that when I click on it, a second tray icon is appearing on the taskbar. Both tray icons have the same menu and both are working. If I click on show again another window with Form1 pops up, and there are three tray icons, and so on....
Can someone help me?

It's because you are creating a new Form1 everytime.
Form1 form1 = new Form1();
You don't want to create a new Form1, you want to show the old one. Give Form2 a reference to your first form1 (call it theMainform1 for example). And then instead of
Form1 form1 = new Form1();
form1.Show();
You want to do
theMainform1.Show();
So you would have:
this.Hide();
Form2 form2 = new Form2();
form2.theMainform1 = this;
form2.Show();

The issue is that you are making a new instance of Form1. This creates a brand new window instead of reviving your old one.
Form1 form1 = new Form1();
form1.Show();
You need to have Form2 reference the original instance of Form1. You can make a constructor to pass in a self reference that would look like
Form2 form2 = new Form2(this);

You can prevent windows from having icons in the taskbar by settings ShowInTaskbar to false on the form. However, the other answers are correct when they say you're creating a new form over and over.
Why would you want two forms to show at the same time? Should they both be on the screen at the same time and be active at the same time? If so, you might try a MDI interface. http://en.wikipedia.org/wiki/Multiple_document_interface
It's possible in WinForms, but I think Microsoft is moving away from them in WPF.

Here is a working code in case that someone is looking for it:
Form1:
/* Hiding Form1 and showing Form2 */
private void btnHideForm1_Click(object sender, EventArgs e)
{
Form mod = new Form2();
mod.Owner = this;
mod.Show();
this.Hide();
}
Form2:
/* Hiding Form2 and showing Form1 */
private void btnHideForm2_Click(object sender, EventArgs e)
{
this.Owner.Show();
this.Close();
}
Thanks for your help guys!!!
I LOVE YOU!!!

Get Application first run Form

I have two Forms in a project.
When my application runs, Form1 is opened. After that I am opening Form2.
How can I access Form1 from Form2 with reflection?

Why would you want to use reflection for this?
When you create the second form, just pass in a reference to the first one:
// I assume it's code within Form1 which opens Form2
Form2 form2 = new Form2(this);
form2.Show();
That's assuming you're happy to add a constructor with Form2 as a parameter. Alternatively, make it a property in Form2:
Form2 form2 = new Form2 { Form1 = this };
form2.Show();

if you open just one instance of form2 you can do this too: Form2 f2 = Application.OpenForms["Form2"];

Better way to Display forms one on top of another

I have a Main form.I want to launch another form from it and launch another from the launched form.I want to ensure that the Main form is not editable when sub forms are displayed so i use showdialog()
Mainform>(Showdialog)>form1>(showDialog+dispose)>form2(dispose)>Mainform
From Mainform i call form2.ShowDialog() then from form2 i use the following code to launch another form
this.visible=false;
form3.showdialog();
this.dispose();
But there are some problems in this.Is there a better way to achieve what im looking for?
edit:more description
I have a Main form,User clicks a Button on Mainform>Form1 is lauched>User clicks a Button in Form1>Form 2 is lauched(diposing/hiding form1) after form2 is closed Mainform should be brought to front and made editable,until then all other forms should be on top of Mainform and Mainform should be un-editable

The problem is that you have to specify the MainForm as the parent for (both) form2 and form3. When you use the overload of ShowDialog that has no parameters, WinForms uses the active form as the parent, so form3's parent becomes form2 automatically. You are then trying to close/dispose form2 causing form3 to become orphaned.
There are several options for getting the reference to MainForm, but the simplest is to use:
form2/3.ShowDialog(Application.OpenForms["MainForm"]);
Assuming that you have set the Name property on MainForm to "MainForm".

In your code, this.dispose() is executed only after form3 is closed. What i think you want is to close form2 after form3 was closed, so you can call this.Close() instead of this.Dispose().
this.visible=false;
form3.showdialog();
this.Close();
Or maybe, after form3 is shown up you dont need form2 any more. That meas:
this.visible=false;
//show instead of showdialog so it wont wait until form3 is closed
form3.show();
this.Close();

It looks like you are trying to implement something like a wizard. The best solution would be to launch all the child forms sequentially, in the main form.
If you need to pass data along the sequence, you should pass it from each dialog to the main form, which then passes it to the next dialog.

MainForm:
Form1 f = new Form1();
if (f.ShowDialog(this) == DialogResult.OK) {
Form2 f2 = new Form2();
f2.ShowDialog(this);
}
Form1 (button click which will open the form 2):
button1_click(object sender, EvengArgs e) {
this.DialogResult = DialogResult.OK;
Close();
}