We Keep Coding

Error with showing a new form in C#

I need your help please
I have created a form and insert a timer & progress bar into it
when the value of progress bar reach to 100% I want to close this form and open the main form of my program
I write this code but when I Run the program it show this error :
( Form that is already displayed modally cannot be displayed as a modal dialog box. Close the form before calling showDialog.)
How I can resolve this problem
Form1 MainForm = new Form1();
public Welcome_window()
{
InitializeComponent();
timer1.Start();
}
private void timer1_Tick(object sender, EventArgs e)
{
progressBar1.Increment(10);
if (progressBar1.Value == 100)
{
this.Visible = false;
MainForm.Visible = false;
MainForm.ShowDialog();
this.Close();
}
}
}

I think the problem is that you don't stop the timer, so the tick event will be fired even if the progress already reached 100%.
Form1 MainForm = new Form1();
public Welcome_window()
{
InitializeComponent();
timer1.Start();
}
private void timer1_Tick(object sender, EventArgs e)
{
progressBar1.Increment(10);
if (progressBar1.Value == 100)
{
timer1.Stop();
this.Visible = false;
MainForm.ShowDialog();
this.Close();
}
}

Nothing happens when I close maximized MDI child form

I have Form1 with 2 radio buttons (rb1 and rb2) and one ordinary button (btn). When I click on btn I should open Form2, as MDI child of Form1 if rb1 is checked, or as ordinary Dialog if rb2 is checked. Also, there can only be one Form2 opened at any moment.
This is my code:
public partial class Form1 : Form
{
Form2 f2;
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
if (f2 != null)
{
MessageBox.Show("Close form!");
return;
}
f2 = new Form2();
if (radioButton1.Checked == true)
{
this.IsMdiContainer = true;
f2.MdiParent = this;
f2.Show();
}
else
{
f2.Show();
}
f2.FormClosed += f2_FormClosed;
}
void f2_FormClosed(object sender, FormClosedEventArgs e)
{
this.IsMdiContainer = false;
f2 = null;
}
}
Everything works as it should except when I maximize Form2 as MDI child and then close it. After that screen stays the same (as I even didn't closed Form2) but I am able to open new Form2, and then Form1's title is "Form1 - [Form2]", and if I repeat the process it would be "Form1 - [Form2] - [Form2]", etc.
I figured out that I my f2_FormClosed method should be
void f2_FormClosed(object sender, FormClosedEventArgs e)
{
f2.Hide(); // <<<<<<<<-----------NEW
this.IsMdiContainer = false;
f2 = null;
}
but I don't know why; Form2 should be closed, I don't know why should I have to hide it?!
Thanks!

I agree with Hans, switching IsMdiContainer at run-time is wonky and is likely to produce other side-effects you haven't seen yet.
Seriously consider a different design for your app.
With that in mind, here's probably the stupidest hack I'll post all day:
public partial class Form1 : Form
{
Form2 f2;
System.Windows.Forms.Timer tmr = new System.Windows.Forms.Timer();
public Form1()
{
InitializeComponent();
tmr.Interval = 100;
tmr.Enabled = false;
tmr.Tick += delegate (object sender, EventArgs e) {
tmr.Stop();
this.IsMdiContainer = false;
};
}
private void button1_Click(object sender, EventArgs e)
{
if (f2 != null)
{
MessageBox.Show("Close form!");
return;
}
f2 = new Form2();
f2.FormClosed += delegate(object sender2, FormClosedEventArgs e2) {
f2 = null;
};
if (radioButton1.Checked == true)
{
this.IsMdiContainer = true;
f2.FormClosed += delegate(object sender3, FormClosedEventArgs e3) {
tmr.Start();
};
f2.MdiParent = this;
}
f2.Show();
}
}
*I originally tried Invoking the call to change IsMdiContainer but that didn't work, so I switched to the Timer. Stupidity that works. Use this solution with caution...

C# close all forms

I have a problem with my program.
I have 3 Forms: First one opens a second form. Second one opens a third form or returns to the first form. The third form can open the first or second forms.
This is how I open a second Form:
private void Open_second_form()
{
Form2 myForm = new Form2(Type_person);
this.Hide();
myForm.ShowDialog();
this.Close();
}
The rest of the forms I open exactly the same.
Here is a code for how I close forms. Every Form has this method:
private void Form1_FormClosing(object sender, FormClosingEventArgs e)
{
if (MessageBox.Show("Exit or no?",
"My First Application",
MessageBoxButtons.YesNo,
MessageBoxIcon.Information) == DialogResult.No)
{
this.Close();
Environment.Exit(1);
}
}
When I open the third form I get 3 MessagesBoxes. If I open first form I got only 1 MessageBox.
I want to close all forms while getting only one MessageBox.
I tried a lot of solutions but none worked.
I tried Application.exit();
Please help me :)

Your confirmation message is funny and result is unobvious =D
There are 2 solutions possible to your issue.
1) If user chooses to close application - don't display confirmation anymore
private static bool _exiting;
private void Form1_FormClosing(object sender, FormClosingEventArgs e)
{
if (!_exiting && MessageBox.Show("Are you sure want to exit?",
"My First Application",
MessageBoxButtons.OkCancel,
MessageBoxIcon.Information) == DialogResult.Ok)
{
_exiting = true;
// this.Close(); // you don't need that, it's already closing
Environment.Exit(1);
}
}
2) use CloseReason to confirm only user actions
private void Form1_FormClosing(object sender, FormClosingEventArgs e)
{
if (e.CloseReason == CloseReason.UserClosing)
{
if(MessageBox.Show("Are you sure want to exit?",
"My First Application",
MessageBoxButtons.OkCancel,
MessageBoxIcon.Information) == DialogResult.Ok)
Environment.Exit(1);
else
e.Cancel = true; // to don't close form is user change his mind
}
}

I'm using this snippet always on my menu form. I hope this helps.
for (int i = Application.OpenForms.Count - 1; i >= 0; i--)
{
if (Application.OpenForms[i].Name != "Menu")
Application.OpenForms[i].Close();
}

Call the Environment.Exit(0); method
private void btnExit_Click(object sender, EventArgs e)
{
Environment.Exit(0);
}

For the form 1, considering the button is called btnForm1To2
private void btnForm1To2_Click(object sender, EventArgs e)
{
using (Form2 myForm = new Form2(Type_person))
{
this.Hide();
myForm.ShowDialog();
this.Show();
}
}
Inside FORM 2, considering the button to page 1 is btnForm2To1, and the button to page 3 is btnForm2To3
private void btnForm2To1_Click(object sender, EventArgs e)
{
// Just close the form 2, as it was called from form 1, this form will be displayed
this.DialogResult = DialogResult.OK;
}
private void btnForm2To3_Click(object sender, EventArgs e)
{
using (Form3 myForm = new Form3(...))
{
this.Hide();
DialogResult form3result = myForm.ShowDialog();
// Now handle the results from form 3 to navigate to form 2 or 1
// In my example, form 3 replies Abort to go back to 1. On cancel (or other dialog results) we will stay on form 2
if (DialogResult.Abort.Equals(form3result))
{
this.DialogResult = DialogResult.OK;
}
else
{
this.Show();
}
}
}
Inside FORM 3, considering the button to page 1 is btnForm3To1, and the button to page 2 is btnForm3To2
private void btnForm3To1_Click(object sender, EventArgs e)
{
this.DialogResult = DialogResult.Abort;
}
private void btnForm3To2_Click(object sender, EventArgs e)
{
this.DialogResult = DialogResult.OK;
}
And if you want that the user confirms his click with a MessageBox, just place them before the this.DialogResult in each button click method

Note that this way is faster than Environment.Exit(0);.
WPF
Set your Window Closing Event in (###Window.xaml)
Closing="Window_Closing"
and then in (###WWindow.xaml.cs)
private void Window_Closing(object sender, System.ComponentModel.CancelEventArgs e)
{
foreach(Window w in Application.Current.Windows)
if(w != this)
w.Close();
}
Windows Form
Go to Form#.cs [Design] >> Form# Properties
Then click on Events
And double-click on FormClosing
In (Form#.cs)
If you're closing from your Startup form, it's better to use this code.
private void Form1_FormClosing(object sender, FormClosingEventArgs e)
{
for(int i = 1; i < Application.OpenForms.Count; ++i)
Application.OpenForms[i].Close();
}
else (Works for any form)
private void Form7_FormClosing(object sender, FormClosingEventArgs e)
{
for(int i = 0; i < Application.OpenForms.Count; ++i)
if(Application.OpenForms[i] != this)
Application.OpenForms[i].Close();
}

If you'd like to close all forms (particularly dialog windows) except for the main form, I've found this function to be useful.
public static void closeAll()
{
FormCollection fc = Application.OpenForms;
if (fc.Count > 1)
{
for (int i = (fc.Count); i > 1; i--)
{
Form selectedForm = Application.OpenForms[i - 1];
selectedForm.Close();
}
}
}

Application.Current.Shutdown();
That's what I use whenever I need to close an entire application. It closes all forms, so great for closing everything. I don't know if this is exactly what you want.

I used this and it worked for me:
for (int i = Application.OpenForms.Count - 1; i >= 0; i--)
{
if (Application.OpenForms[i].Name != "FRM01"&&Application.OpenForms[i].Name!= "FRM02")
Application.OpenForms[i].Close();
}
Those "FRM" things are the names of my forms, what you can do there is to add as many forms as you like to have as "parents". So each time it iterates it will check if the names of those forms are the ones you wish to keep.
I recommend you add that 'for' into your form closing handler if you wish to control an user from closing parent forms and child forms to still be there.
It is always better to use MDI forms (MDI link here)

a problem with bringing form in front

I a form than I can open it also by pressing F1 and clicking so in this code first I check if the form has been opened or not.if was opened I just want to bring it to front. my problem is it wont bring in front by pressing F1 or clicking ToolStripMenuItem if i open it befor.
private void Form1_KeyDown(object sender, KeyEventArgs e)
{
Form2 form2;
form2 = new Form2();
if (e.KeyCode.ToString() == "F1" && Application.OpenForms.OfType<Form2>().Count() <= 0)
// the user pressed the F1 key
form2.Show();
else
{
form2.TopMost = true;
form2.BringToFront();
form2.Focus();
form2.TopMost = false;
}
}
private void ToolStripMenuItem_Click(object sender, EventArgs e)
{
Form2 form2;
form2 = new Form2();
if (Application.OpenForms.OfType<Form2>().Count() <= 0)
form2.Show();
else
{
form2.TopMost = true;
form2.BringToFront();
form2.Focus();
form2.TopMost = false;
}
}

It doesn't work because you forgot to call form2.Show() in the else clause. The code is wrong, you don't want to create a new instance of Form2 if one already exists. Also, there's a bug in Winforms that makes Application.OpenForms lose track of form instances.
Best thing to do is to explicitly keep track of the lifetime of the form with its FormClosed event handler so you don't have to find it back later:
Form2 form2;
private void showForm2() {
if (form2 == null) {
form2 = new Form2();
form2.FormClosed += delegate { form2 = null; };
}
form2.Show();
form2.Focus();
}
Call showForm2 from your event handlers.

Just call form2.Show() and BringToFront. You should declare the `new Form2()' once though.
private Form2 form2;
private void ShowForm2()
{
if (form2 == null)
{
form2 = new Form2();
form2.FormClosed += delegate { form2 = null; };
}
form2.Show();
form2.BringToFront();
}
private void Form1_KeyDown(object sender, KeyEventArgs e)
{
ShowForm2();
}
private void ToolStripMenuItem_Click(object sender, EventArgs e)
{
ShowForm2();
}
From Hans' code, I added the delegate to handle null problem on closing the form and opening it again.

how to prevent opening a form multiple times in c#

i have created an application in which a menustrip is present in which 2 buttons are there, one for ADD, and another for UPDATE & both controls are in a single form, means a button of add & update is there in a single form, whenever i press add button in menustrip update button will be disabled, and when i press update on menustrip the add button will disable. how to do this? i m doing this by show method but that form is opening multiple times using show().
private void addRecordsToolStripMenuItem_Click(object sender, EventArgs e)
{
Form1 f2 = new Form1();
f2.MdiParent = this;
f2.Show();
f2.button1.Enabled = true;
}
private void updateRecordsToolStripMenuItem_Click(object sender, EventArgs e)
{
Form1 f2 = new Form1();
f2.MdiParent = this;
f2.Show();
f2.button2.Enabled = true;
f2.button1.Enabled = false;
}

you simply have to use a single form in this case. try using the singleton approach -
http://hashfactor.wordpress.com/2009/03/31/c-winforms-create-a-single-instance-form/

try using .ShowDialog() instead .Show() and no other form will be able to be clicked on until that one closes.

To do that you'll need to have an instance of that Form outside of those methods that you dismply show if the Form has already been created, or create and show it if it has not (this is the singleton pattern). Here's an example:
Form1 f2 = null;
private void addRecordsToolStripMenuItem_Click(object sender, EventArgs e)
{
if (f2 == null)
{
f2 = new Form1();
f2.MdiParent = this;
f2.button1.Enabled = true;
}
f2.Show();
}
private void updateRecordsToolStripMenuItem_Click(object sender, EventArgs e)
{
if (f2 == null)
{
f2.MdiParent = this;
f2.button2.Enabled = true;
f2.button1.Enabled = false;
}
f2.Show();
}
One question on your disabling of the menu items though, how do you plan on re-enabling them after they have been disabled?

just try to check that form is already opened or not by using its Text Property.. if it is opened just focus on that form other wise show that form as normally
private void button1_Click(object sender, EventArgs e)
{
bool IsOpen = false;
foreach (Form f in Application.OpenForms)
{
if (f.Text == "Form1")
{
IsOpen = true;
f.Focus();
break;
}
}
if (IsOpen == false)
{
Form f1 = new Form1();
f1.Show();
}
}
Try This Guys Its Simple