Windows Form project restart button click then comes error?

Question

My Windows From project debugs correctly. But after I will click "Sign Out" button (It's used Application.Restart() function) then comes this error. Please give me some advice.

Amit Gupta · Answer

In my very earlier of windows application development phase, I found that windows form might not be able to perform some of the simple task. I tried all the stuffs to make the form invisible to user, you may also try to achieve the task. private void Logout_Click() { this .Close(); ShowInTaskbar = false ; Visible= false ; Opacity=0; new LoginForm().ShowDialog(); } You can use one/all to make this work. Clean & Rebuild your project, sometimes this also works too..

sampath tharanga · Answer

This error comes again. It comes Exit button click. so, I guess this error come using Application.Exit() and Application.Restart() function use. I can't understand where a problem.

sampath tharanga · Answer

yeah, I sure, This my code parts. private void btnLogout_Click(object sender, EventArgs e) { Close(); new LoginForm().ShowDialog(); } static void Main() { Application.EnableVisualStyles(); Application.SetCompatibleTextRenderingDefault(false); Application.Run(new LoginForm()); } private void btnSignIn_Click(object sender, EventArgs e) { ....... this.Hide(); MainForm ObjMn = new MainForm(); ObjMn.ShowDialog(); this.Close(); } But where my mistake. Can't find it, brother. But, this working private void btnLogout_Click(object sender, EventArgs e) { Close(); new LoginForm().ShowDialog(); } But, always hide use not a good way. so...

sampath tharanga · Answer

It's working. Brother, then it does not have any side effect or something. Thanks a lot.

Amit Gupta · Answer

Have you applied all three snippet that I had posted? It will definitely work, if it doesn't then you might be doing some mistakes. Its better to show your code, what you had done.

sampath tharanga · Answer

yeah. Just like my code. different restart() function use for sign out button. Now I change it private void Logout_Click() { Close(); new LoginForm().ShowDialog(); } just like this. but does not working. sign oclickutton clik then open LoginForm. But just not close the MainForm. Please help me.

Amit Gupta · Answer

Hey You are in wrong approach, if you want to signout the your form will close the main form and open the login window. // your main method static void Main() { ..... Application.Run ( new LoginForm()); } // Login button clicked in Login Window private void Login_click() { Hide(); // start your main form new MainForm().ShowDialog(); } // your logout button on Main window private void Logout_Click() { Close(); new LoginForm().ShowDialog(); } Just that. Good Luck

Nitin Sontakke · Answer

Aha! Understood. However, the restart is not for what you expect it to be. Please refer to the documentation of the .Restart() method online. I have never done what you are describing. Typically this is not done in Windows Forms Applications at all. I think you are trying to duplicate as we typically do on the web-sites. In Windows Forms Application, we typically just exit it. User can always just double click on the icon and start it again and the login form appears. In web scenario, we have to do it so as the session on the server side ends. Here there is no such thing. Hope you get the point.

sampath tharanga · Answer

Now I'm using the "try and catch". Then solve it. And Not comes message box. I can't understand this reason. What is the reason for come previous error? static void Main() { try { Application.EnableVisualStyles(); Application.SetCompatibleTextRenderingDefault(false); Application.Run(new LoginForm()); } catch(Exception ex) { MessageBox.Show(ex.Message); } }

sampath tharanga · Answer

Because It's only "Sign Out". Not the exit. So, Click the sign out button then restart the application and open the "login" form.

Nitin Sontakke · Answer

I hope you expect to close the application when 'Sign Out' button is clicked. Why are you calling Restart() is not clear to me. Please use Application.Exit(). https://msdn.microsoft.com/en-us/library/system.windows.forms.application.exit(v=vs.110).aspx