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How to Open another Form without closing the whole Application c#

I have Forms (Admin, login, user). There is no error when accessing user form, it is simply opening it without user and pass.
The error I encountered is when I login and Admin form is present, when I click in my toolstripbutton the whole application exits.
This is the method I applied in every toolstripbutton:
public void CloseAllActiveForms()
{
List<Form> openForms = new List<Form>();
foreach (Form f in Application.OpenForms)
openForms.Add(f);
foreach (Form f in openForms)
{
if (f.Name != "FrmAdmin")
{
f.Close();
}
}
}
And this is the code in every toolstipbutton:
private void toolStripButton4_Click(object sender, EventArgs e)
{
CloseAllActiveForms();
FrmDashboard objFORM = new FrmDashboard();
objFORM.MdiParent = this;
objFORM.TopLevel = false;
objFORM.FormBorderStyle = FormBorderStyle.None;
objFORM.Dock = DockStyle.Fill;
pnlMain.Controls.Add(objFORM);
objFORM.Show();
}
This is design sample.
Every click in toolstripbutton supposed to go in pnlMain, but the problem occurs that after logging in and click one of the toolstripbutton it quits the whole windows application.
I tried to search about these and I found almost the same as my problem but the solution I think is not for my problem because I think it is for two forms only and the main form will be put in program.cs, but i have 2 and I think but not sure that my main form is the login. Please enlighten me.
Thank you

Because of this line in your program.cs
Application.Run(new FrmLogin());
the form FrmLogin is the main form of your application.
When the main form of an application is closed, the application is terminated. I suppose that after Login, you do not close but only hide the login form.
However, your code in CloseAllActiveForms will also close the hidden login form.
A quick solution is probably to skip all hidden forms in CloseAllActiveForms or explicitly skip FrmLogin in CloseAllActiveForms.
public void CloseAllActiveForms()
{
List<Form> openForms = new List<Form>();
foreach (Form f in Application.OpenForms)
openForms.Add(f);
foreach (Form f in openForms)
{
if (f.Visible && !(f is FrmAdmin) && !(f is FrmLogin))
{
f.Close();
}
}
}
There are alternative cleaner solutions:
Show the login form from the main form. See how to show login form on front & Mainform Behind Login Form?
Pass an ApplicationContext instead of a Form to Application.Run. This allows you to have more control over when the application is closed. See Windows Forms: create the main application after login, which form to run?

Creating new pages in Visual Studio using C# Windows Forms

I am trying to write a Windows Forms application using C#, however I have never done this before. I have successfully created a login page which takes the user to a home page where 4 buttons are displayed. I am trying to find code to put inside each button that will take the user to a different page.

In windows forms application, you have to call Show method on the Form which you want to show.
Let's say on Button1 click you want to show Form2, then below code will bring Form 2 up on the screen.
private void button1_Click_1(object sender, EventArgs e)
{
Form2 obj = new Form2();
if (obj == null)
{
obj.Parent = this;
}
obj.Show();
this.Hide();
}

You need to create, initialize and show the other forms.
frmSecond frm = new frmSecond();//You should call any other constructor, may be with some parameters
frm.Text ="I wanted to change the title";//Optional: You can change any property value if you need
frm.Show();
If you want to show the form, collect some information from user and return some values then you can use ShowDialog method instead of Show method
if(frm.ShowDialog(this) == DialogResult.OK){
var myVar = frm.ReturnObject;
...
}

You can use below approach for one parent window and new child window.
When you press the button new form opens:
private void button1_Click(object sender, EventArgs e)
{
Form2 frm2 = new Form2();
{
frm2.ShowDialog();
}
}

how to open a form from another form in c#

i am writing a code for serial key registration.
if the serial key entered by the user is correct then anothr form must open and the present form must close.
please go thought the code.
namespace ExtTrigger
{
public partial class Activation : Form
{
public Activation()
{
InitializeComponent();
}
private void ActivateButton_Click(object sender, EventArgs e)
{
String key;
key = string.Concat(textBox1.Text,textBox2.Text,textBox3.Text,textBox4.Text);
if (key == "1234123412341234")
{
Properties.Settings.Default.Registered = true;
MessageBox.Show("hurray", "", MessageBoxButtons.OK);
Form1 f1= new Form1();
f1.ShowDialog();
this.Close();
}
else
MessageBox.Show("No Match", "", MessageBoxButtons.OK);
}
private void Activation_Load(object sender, EventArgs e)
{
}
}
my problem is: On clicking on ActivateBotton, Form1 opens but the present form doesnot close.
i have read in few threads that in VB we can change the property: ShutdownMode.
How can we do that in c#?

f1.ShowDialog(); blocks the call, it doesn't go to the next line until that new form has been closed.
An option would be to use:
f1.Show();
Show doesn't block the call, it passes to the next statement. It will not wait for the new form to be closed.

since you have show the second form as f1.ShowDialog() so first one remain open untile second one close, try this
Form1 f1= new Form1();
f1.Show();
this.Close();

The following code should do the trick:
using(Form1 f1 = new Form1())
{
this.Hide();
DialogResult result = f1.ShowDialog();
if(result == DialogResult.OK)
{
this.Show();
}
}
You create your new form within the using-block, then you hide your main form(or the form you are in at the moment) create a DialogResult that gets set by the newly opened form and open this form. Now you can set the results you want to check for inside of your new form and if everything went well inside of you new form you set the DialogResult to OK via:
this.DialogResult = DialogResult.OK;
Now back in our first form you check for the DialogResult and if it is okay you show your main form again. If it was not okay you could just reopen the 2nd form and let the user try again.

Opening a new form is very simple, but the way you do it really depends on your need.
Case 1: I would like to freeze/ block the calling form on secondary form call
In this case you should be using secondaryFormObj.ShowDialog();
Of course, when using this technique your called form, which now acts as a dialog, should "return" an answer to its caller parent on closure.
For example:
private void SecondaryForm_Closing(object sender, System.ComponentModel.CancelEventArgs e)
{
// Just a dummy code example.
// Always returns Yes result on form closure
this.DialogResult = System.Windows.Forms.DialogResult.Yes;
}
For more help and examples on this manner you can use MSDN:
DialogResult - MSDN
Case 2: I would like to have both forms responding at the same time
In this case you basically need to call secondaryFormObj.Show();
If you want the caller form to be hidden on secondary form call, just
invoke this.Hide();
after the call to secondaryFormObj.Show(); in the caller class.
You can also close the caller form using this.Close(); as long as the caller form is not the application's main form.
... And remember
Always make sure you initialized the secondary form object before
invoking it with either secondaryFormObj.Show(); or
secondaryFormObj.ShowDialog();
Initializing a form is done the same way like every typical object using the
new operator.
For example: secondaryFormObj = new Form();
Hopes this helps. Happy coding!

Keep form on top and open and close other forms

I'm new to C# and could use some help.
What I have so far is a set of 8 windows forms I have created in C#, these forms have basic things like text boxes, labels, radio buttons, etc. Now that I have completed making all of these forms I want to have one additional form (called the Selector form) I can use to select one of the other 8 forms. At any given time, I want the Selector form to be on top of other windows and it will have 8 radio buttons (or regular buttons, doesn't matter). When one of the buttons is clicked, the current form (not the Selector form) should disappear and a new form should appear. The name of the button will be the name of the new form that appears.
I have seen a few examples and here is the code I have so far:
void Button1Click(object sender, EventArgs e)
{
//this.Hide();
var form1 = new CASII();
form1.Closed += (sender1, args) => this.Close();
form1.Show();
}
void Button2Click(object sender, EventArgs e)
{
// this.Hide();
var form2 = new CCARAdmin();
form2.Closed += (sender1, args) => this.Close();
form2.Show();
//Application.Run(new CCARAdmin());
}
Problem I am having is I don't want to hide the Selector form, which this does, and I don't know how to identify the other form that is open to close it and then open a different form.
From starting the program the logic would be like this:
Show Selector form
When a button is clicked on the Selector form, keep the Selector form on top and show the other form with the name of the button.
When a different button is clicked on the Selector form, close the previous form that was open (not the Selector form) and open the new form corresponding to the name of the button. Keep the Selector form on top.
When the Selector form is close, application stops.

Problem I am having is I don't want to hide the Selector form, which
this does, and I don't know how to identify the other form that is
open to close it and then open a different form.
Set Selector form TopMost to True to make it always on top. Or
you can use BringToFront after opening a new form
to know other forms that are open check this answer. Or you can define each From as a field in the Selector form, and check that.

selectorForm.TopMost = true ( this will help to keep the selector form always on top).
Create a form variable in your selector form to keep the reference of your currently opened form.
Sample code for 1 button click :
Form frm = null;
void Button1Click(object sender, EventArgs e)
{
//this.Hide();
var form1 = new CASII();
if (frm == null)
{
frm = form1;
}
else
{
frm.Close();
}
form1.Show();
this.TopMost = true;
frm = form1;
}

I resolved this by setting TopMost to true and then using the following code under each one of the buttons:
for (int i = Application.OpenForms.Count - 1; i >= 0; i--)
{
if (Application.OpenForms[i].Name != "FormSelector")
Application.OpenForms[i].Close();
}
var form = new TRAC();
if (radioButton9.Checked == true)
{
form.Show();
}

Close child form if already open

How can I go about closing a Child form if it's already open? And if it isn't already open, then open it?
Thank you
I already have this code, which doesn't work, obviously:
Form ibx = new inbox();
if(ibx.Visible)
ibx.Dispose();
else
ibx.Show();
All that the above does is creates a new form whether or not it's already open.

private Form frm;
public void ToggleForm() {
if(frm == null) {
frm = new Form();
frm.Show();
}
else {
frm.Close();
frm = null;
}
}

If what you want to achieve is opening a form only if it isn't already open, then bring it on front when it is, then the following code might help.
private void tsmiMenuOption_Click(object sender, EventArgs e) {
// Assuming this method is part of an MDI form.
foreach(Form child in this.MdiChildren)
if (child.Name == MyForm.Name) {
child.BringToFront();
return;
}
MyForm f = new MyForm();
f.MdiParent = this;
f.Show();
}
So, this is untested pseudo-c#-code that verifies if MyForm is already opened and contained in the MdiContainer.Children. If it is, then it brings this form (MyForm) to the front. If it is not, then it simply instantiate and opens it.
Is this about what you want?

In your main form, when you create the first childform, keep a reference to it. Currently you're creating a new childform every time you run that code and then you check if the newly created form is visible.
Also be aware that your subject talk about opening and closing but your code seems to just be dealing with hiding and showing.
Carra's code is a good sample how to do it, but be careful if the childform can be closed from anywhere else.

You need to keep a reference to ibx. Your code creates a new inbox everytime it's run.

Module Module1
Public Function InstanceNewForm(ByRef ParentForm As Form, ByRef Childform As Form) As Boolean
Dim bOpen As Boolean = False
Dim frm As Form
For Each frm In ParentForm.MdiChildren
If Childform.Name = frm.Name Then
Childform.Focus()
bOpen = True
Exit For
End If
Next
If Not bOpen Then
With Childform
.StartPosition = FormStartPosition.CenterScreen
.MdiParent = Parentform
.Show()
End With
End If
frm = Nothing
Return bOpen
End Function
End Module
The above code will check to see if a mdi form is already loaded in the parent container and if its already active will set the focus to it. Otherwise it will create the mdi form.
Just call it from anywhere the mdi form needs to be loaded. ex: call InstanceNewForm(me,form2)
Works like a charm everytime!!