My C# Winform application encounters the situation where it cannot access a disposed object. The disposed object is a form (frmQuiz) that is opened from a button on the login form.
The Situation:
My application typically has two or three forms open at the same time. The Program.cs file runs form frmLoginBackground, which is just a semi-transparent background that covers the computer screen. The load event for this form opens the second form, frmLogin, which includes a button that opens frmQuiz, which is a simple form with a few math questions on it.
The code in frmLogin that opens frmQuiz looks like this:
private void btnTakeQuizNow_Click(object sender, EventArgs e)
{
frmQuiz quiz = new frmQuiz();
quiz.TakeQuizNow("take_quiz_now", Convert.ToInt32(comboQuizMeNow.SelectedValue)); //Pass the form a quiz id number.
quiz.Show();
}
When frmQuiz opens both it and frmLogin are open and accessible.
The frmLogin also contains a password control that opens the administration form by first opening frmSplash, which is a "Please Wait..." splash form based on a timer. The timer Tick event launches frmAdmin, which is the administration form. The code in frmLogin looks like this:
private void btnPasswordSubmit_Click(object sender, EventArgs e)
{
//Password verification code snipped.
frmSplash objSplash = new frmSplash();
objSplash.Show();
//this.Hide();
this.Close();
}
And the code in frmSplash looks like this:
private void timer1_Tick(object sender, EventArgs e)
{
frmAdmin objfrmAdmin = new frmAdmin ();
objfrmAdmin.Show();
this.Close();
}
When frmAdmin opens then frmLogin is no longer accessible; however, frmAdmin contains a 'Return to Login Screen' button with code like this:
private void btnReturnToLogin_Click(object sender, EventArgs e)
{
exitWarnings("return_to_login");
}
private void exitWarnings(string action)
{
//Warning message code snipped.
if (action == "return_to_login")
{
frmLogin objLogin = new frmLogin();
objLogin.Show();
}
}
The frmLoginBackground remains open until the application exits.
The Problem:
Everything works fine when frmLogin first opens and the button is clicked to open frmQuiz. The quiz form opens a runs fine. However, after logging into the administration form (which closes or hides the login form) and then clicking the 'Return to Login Screen' link, then, after frmLogin reappears, the object disposed exception occurs when clicking the button to open frmQuiz. Visual Studio highlights in yellow the "quiz.Show();" line of code. The exception occurs regardless of weather I use "this.Close();" or "this.Hide();" in the btnPasswordSubmit_Click event.
Can anyone suggest a solution that allows me to open frmQuiz after returning to frmLogin from frmAdmin.
Cheers, Frederick
Since you create a new instance for quizz just before the quizz.Show() it cannot be quizz itself that throws the exception.
Take a good look at the constructor and FormCreate event of frmQuiz. It looks like that is where the dead horse is being kicked.
Related
So, I'm making a payroll management system as a hobby project to help my resume and general knowledge of c#. So, I'm making a UI and I can open a new window just fine with this code:
private void button1_Click(object sender, EventArgs e)
{
CreateAdminAcct createAcct = new CreateAdminAcct();
createAcct.StartPosition = FormStartPosition.CenterScreen;
createAcct.Show();
this.Hide();
}
however, I don't know the event to check when the little red "x" button is clicked, because when that button is clicked, I want to go back to the main screen because I hide the main screen when that button is clicked, and when i click the red "x" on the screen that just opened, it closes, but the application continues to run in the background.
If there is some better way to manage multiple menus, I'm open to suggestions, however, this is what I've found easiest.
Thanks in advance
I second Robert Harvey's suggestion; this gives the user the reassurance tha tht emain window is still open/ nothing got lost, but it's unreachably "behind" the CreateAdminAcct form while the CreateAdminAcct form is open
private void button1_Click(object sender, EventArgs e)
{
CreateAdminAcct createAcct = new CreateAdminAcct();
createAcct.StartPosition = FormStartPosition.CenterScreen;
createAcct.ShowDialog();
//do any code here that needs to access createAcct before it's lost
MessageBox.Show(createAcct.NewAdmin.Name);
}
If you really do want to hide your main form, pass the main form itself to createAcct, and make it createAcct's job to re-open the main form when it is closing
private void button1_Click(object sender, EventArgs e)
{
CreateAdminAcct createAcct = new CreateAdminAcct(this); //note passing this form to constructor
createAcct.StartPosition = FormStartPosition.CenterScreen;
createAcct.Show();
}
class CreateAcctForm : Form{
private Form _showWhenClosing;
CreateAcctForm(Form revertTo){
InitializeComponent();
_showWhenClosing = revertTo;
}
}
void Form_Closing(object sender, ...){ //event
_showWhenClosing.Show();
}
Side note: please rename your controls after you drop them ona form. code that's stuffed with label57, textbox25 is effectively obfuscated and really wearisome to follow
well I have a funny problem with closing dialog forms.
here is the problem:
I run application and open second form (through menu strip) as showdialog(); and then from second form open third form. When I open third form by button 1 and then close it everything is alright, but when I open the third form by button 2 and then close it, third form will be closed and then it closes the second form also. !!! In second form when I show a messageBox and close it also the second form will be closed.
here is my codes:
open second form from first form codes:
private void settingsToolMenu_Click(object sender, EventArgs e)
{
settingsForm s1 = new settingsForm(this);
s1.ShowDialog();
}
open third form from second by button 1 form codes:
private void addReportButton_Click(object sender, EventArgs e)
{
addReport a1 = new addReport(this);
a1.ShowDialog();
}
open third form from second by button 2 form codes:
private void editReportButton_Click(object sender, EventArgs e)
{
addReport a2 = new addReport(this);
a2.ShowDialog();
}
as you see there is no differences between button 1 and button 2
here is a video from application running.
Not sure what's happening out there, but there should be .Show() method, which runs a window in a different way including closing strategy. Try it out.
Try This
Instead of
addReport a2 = new addReport(this);
a2.ShowDialog();
Use
addReport a2 = new addReport();
a2.ShowDialog(this);
Then on click of Exit / Close button of dialog window
private void BtnExit_Click(object sender, EventArgs e)
{
this.Dispose();
}
Hope this will solve your issue.
I used this code and it worked. I have 3 forms, the first form is opened when running the app, the second form is opened with a button (can be menustrip, doesn't matter), then the third is opened like that too, after closing the third form the second form remains open.
FormN fm = new FormN();
fm.ShowDialog();
Use that piece of code in every method that is called from clicking on a button and it should work fine. Just change the "FormN" for whatever your forms are named. Also, if you need to pass any form's attributes into the next form you can do this:
Code at first form:
public string mytext; //Variable I want to use later, in Form2.
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
mytext = tb1.Text;
Form2 fm = new Form2(this);
fm.ShowDialog();
}
Notice how I save "tb1"'s (TextBox1) value in a variable before calling "fm.ShowDialog();", so I can use the TextBox1 value later inside the Form2.
Code at second form, having main form's variables (such as "mytext" value).
Form1 mfm;
public Form2(Form1 mainfm)
{
InitializeComponent();
mfm = mainfm;
}
public void button2_Click(object sender, EventArgs e)
{
//In this method I use the variable "mytext" wich is a Form1 attribute.
//You can see how I declare it in the first form's code (see above).
textBox1.Text = mfm.mytext;
}
With this you have created an object of your main form ("Form1 mfm;") with all the variables it contained before calling the second form, which can be used for the third form too.
in second form formClosing() event i wrote these codes:
private void settingsForm_FormClosing(object sender, FormClosingEventArgs e)
{
if(e.CloseReason != CloseReason.UserClosing)
{
e.Cancel = true;
}
}
and nothing can close the second form except user!
i'm new to the programming went through few tutorials and sample projects and then started to create my own text based adventure game with some UI.
So what i'd like to achieve with the beginning of my project is, when user launches exe, i'd like to greet them with a username input screen with Start and Exit buttons and then close that form, launch a new form which i'll put in the game's main interface.
So, when i click the "Start" Button, it'll read the username from the textbox, save it to a string, close the form and launch a new form with also using the name screen in the game's main interface.
My question is, How can i link the start button from the below code to a new Form, also closing the current AUJFM_Login form, which will also be able to read the string username.
I have tried few things but after a few attempts, i just left it with the button functions. It's not much but here is the basics of it:
The Greeting screen will be called AUJFM_Login, and the main interface will be called AUJFM.
namespace AUJFM
{
public partial class AUJFM_Login : Form
{
public AUJFM_Login()
{
InitializeComponent();
}
private void btnStart_Click(object sender, EventArgs e)
{
string UserName = nameBox.Text;
}
private void btnExit_Click(object sender, EventArgs e)
{
Application.Exit();
}
}
You can try the code below for the “Start” button click. I added a method to check the user name as it seems like sending an invalid user name to the next form is simply a waste of time. You will have to adjust this method to check for the valid users. Hope this is what you are looking for.
private void btnStart_Click(object sender, EventArgs e) {
string userName = nameBox.Text;
if (ValidUserName(userName)) {
SecondForm nextForm = new SecondForm(userName);
nextForm.Show();
this.Hide();
} else {
// user name not valid
}
}
private bool ValidUserName(String userName) {
// logic to check if user name is valid
return true;
}
Then in the second form constructor, change the signature to accept the user name string.
public SecondForm(string userName) {
InitializeComponent();
textBox1.Text = userName;
}
If you have a form for the main window (Let's call it MainForm),
you can do:
MainForm mainForm = new MainForm();
mainForm.Show();
The main window would then appear.
To close the login form, you could do
this.Hide();
Since closing the form from which the application runs would close the entire application.
I want to close two forms at the same time. I have one main form that starts with program. when user click button, the main form will hide and other form will pop up. on the second form if user click "back to main " button, it should hide second form and show main form. But the problem is if user tries to close the second form it should close the main form as well. How can i close the main form as well
I would just use the Application.Exit() for what is requested by this thread.
Application.Exit();
UPDATE: corrected
I had said this will not call the form closing events but in documentation it does actually call it here is a link to the documentation
http://msdn.microsoft.com/en-us/library/ms157894(v=vs.110).aspx
It was better if you specified what codes you wrote for going back to main form, so I could help you by changing your codes. But now because I don't know how you did it, I have to write codes for both of those tasks.
It can be possible using a Boolean variable to do what you want. Follow bellow codes:
public partial class MainForm : Form
{
//"Click" event of the button that should opens the second form:
private void goToSecondForm_Click(object sender, EventArgs e)
{
Form2 f2 = new Form2(); //Or you can write it out of this method.
this.Hide(); //Hides the main form.
f2.ShowDialog(); // Shows the second form.
this.Show(); // Shows the main form again, after closing the second form using your own button.
}
}
public partial class Form2 : Form
{
bool selfClose = false; //False shows that user closed the second form by default button and true shows that user closed it by your own button.
//"Click" event of the button that should closes just the second form and returns user to the main form:
private void ownCloseButton_Click(object sender, EventArgs e)
{
selfClose = true; //Means user clicked on your own button.
this.Close(); //So the program closes the second form and runs f2_FormClosed method, but because selfClose became true here, happened nothing there and program will go back to goToSecondForm_Click method in the main form and will run this.Show() .
}
//"FormClosed" event of the second form :
//Whether user clicked on your own button or on the default one, this method will run.
private void f2_FormClosed(object sender, FormClosedEventArgs e)
{
if (!selfClose) //It means user didn't click on your own button and both of forms must be closed .
Application.Exit(); //So the program closes all of forms (actually closes the program) and couldn't access to any other commands (including this.Show() in goToSecondForm_Click method).
}
}
As others have said, you need to somehow call .Close() on the main form when your child form is closed. However, as you've pointed out, you don't have a reference to the main form automatically in your child form! That leaves you with a few options.
1. Exit the application immediately.
This is done by calling Application.Exit(); in your child form's "back to main" button's click event handler. It will immediately close all forms, which might simply be what you want.
// .. ChildForm code ..
void OnBackToMainClicked(object sender, EventArgs e)
{
Application.Exit();
}
2. Pass a reference to the main form to the child form.
This is probably the most common way to solve this problem in general. When you create your child form in your main form, you'll need to pass a reference as follows:
// .. MainForm code ..
void OnGoToChildForm(object sender, EventArgs e)
{
var childForm = new ChildForm(this);
childForm.Show();
}
// .. ChildForm code ..
private MainForm mainForm; // This is where the child form will keep a reference to
// the main form that you can use later
public ChildForm(MainForm mainForm)
{
// This is the child form's constructor that we called above,
// and it's where we'll save the reference to the main form
this.mainForm = mainForm;
}
// This also needs to be the event handler for the close event
void OnBackToMainClicked(object sender, EventArgs e)
{
this.Close();
mainForm.Close();
}
3. Add an event handler on the child form's FormClosed event. This is a safe way to solve the problem if you are concerned about keeping your main application logic under the control of the main form. It's similar to the solution suggested by Lamloumi above, but it's all done in the main form's code.
// .. MainForm code ..
void OnGoToChildForm(object sender, EventArgs e)
{
var childForm = new ChildForm(this);
childForm.FormClosed += new FormClosedEventHandler(SecondForm_FormClosed);
childForm.Show();
}
void SecondForm_FormClosed(object sender, EventArgs e)
{
// Perform any final cleanup logic here.
this.Close();
}
Form1 _FirstForm = New Form1();
Form2 _SecondForm = New Form2();
MainForm _MainForm = new MainForm();
_FirstForm.Close();
_SecondForm.Close();
_MainForm.Show();
Normally , in the Home form you have some ting like this :
SecondForm second= new SecondForm ();
second.Show();
this.Hide();
In the SecondForm you must ovveride the event of closure like this :
public class SecondForm :Form{
public SecondForm()
{
InitializeComponent();
this.FormClosed += new FormClosedEventHandler(SecondForm_FormClosed);
}
void SecondForm_FormClosed(object sender, FormClosedEventArgs e)
{
Application.Exit();
}
}
To be sure that your application is closed after you close the form. because the Home from is still active and hidden if you don't.
use this in Form2
private void button1_Click(object sender, EventArgs e)
{
Form1.FromHandle(this.Handle);
}
There Handle are the same now ;
hope this work.
I've looked at all the suggested answers and nothing seems to fit what I'm looking for. I want to call a second form from my main form, hide my main form while the second form is active, and then unhide the main form when the second form closes. Basically I want to "toggle" between the two forms.
So far I have:
In my main form:
private void countClick(object sender, EventArgs e)
{
this.Hide();
subForm myNewForm = new subForm();
myNewForm.ShowDialog();
}
and in my second form I have:
private void totalClick(object sender, EventArgs e)
{
this.Close();
}
How do I get the main form to show?
ShowDialog opens your secondary Form as Modal Dialog, meaning that the MainForm's code execution will stop at that point and your secondary Form will have the focus. so all that you need to do is put a this.Show after your ShowDialog call.
From above link:
You can use this method to display a modal dialog box in your application. When this method is called, the code following it is not executed until after the dialog box is closed.
private void countClick(object sender, EventArgs e)
{
this.Hide();
subForm myNewForm = new subForm();
myNewForm.ShowDialog();
this.Show();
}
Let's say in Form1 you click a Button to show Form2
Form2 frm2 = new Form2();
frm2.Activated += new EventHandler(frm2_Activated); // Handler when the form is activated
frm2.FormClosed += new FormClosedEventHandler(frm2_FormClosed); // Hander when the form is closed
frm2.Show();
Now, this one is when the Form2 is shown or is Activated you hide the calling form, in this case the Form1
private void frm2_Activated(object sender, EventArgs e)
{
this.Hide(); // Hides Form1 but it is till in Memory
}
Then when Form2 is Closed it will Unhide Form1.
private void frm2_FormClosed(object sender, FormClosedEventArgs e)
{
this.Show(); // Unhide Form1
}
This is difficult to do correctly. The issue is that you must avoid having no window at all that can get the focus. The Windows window manager will be forced to find another window to give the focus to. That will be a window of another application. Your window will disappear behind it.
That's already the case in your existing code snippet, you are hiding your main window before showing the dialog. That usually turns out okay, except when the dialog is slow to create. It will definitely happen when the dialog is closed.
So what you need to do is hide your window after you display the dialog and show it again before the dialog closes. That requires tricks. They look like this:
private void countClick(object sender, EventArgs e)
{
this.BeginInvoke(new Action(() => this.Hide()));
using (var dlg = new subForm()) {
dlg.FormClosing += (s, fcea) => { if (!fcea.Cancel) this.Show(); };
if (dlg.ShowDialog() == DialogResult.OK) {
// etc...
}
}
}
The BeginInvoke() call is a trick to get code to run after the ShowDialog() method runs. Thus ensuring your window is hidden after the dialog window is shown. The FormClosing event of the dialog is used to get the window to be visible again just before the dialog closes.
You need to find some way to pass a reference to the main form to the second form click event handler.
You can do this either by setting the form as a member variable of the second form class or pass it via the event arguments.
If you are working in the same namespace, you have the context, using mainform or the name you gave the "main form", try:
mainform.show();