I have a Form1 and another one that I added. Form1 is being run by program.cs at the start. I need to hide Form1 and show options form by the press of a button.
private void submitPassword_Click(object sender, EventArgs e)
{
options optionForm = new options();
optionForm.Show();
}
the above code opens the options form on top, but I need it to replace the current form. how can I achieve this?
Hide current form using this.Close() before showing new one and make sure you are using parameterless Application.Run so program won't close when you close it's main form.
You can use "Hide" and "Show" method :
private void submitPassword_Click(object sender, EventArgs e)
{
options optionForm = new options();
this.Hide();
optionForm.Show();
}
private void submitPassword_Click(object sender, EventArgs e)
{
options optionForm = new options();
optionForm.Show();
this.Hide();
}
Similar solutions where one form calls and acts on another... Such as this one I answered for another. You could do a similar process... pass in your first form to the second... Then show the second... Then, you could HIDE your first form (via this.Hide() ). Then, in your second form, when you click whatever button to select your choice, and need to return back to the first form, you could then use the original form's reference passed INTO the second form to re-Show it, such as in the click on the second form...
this.PreservedForm.Show(); // re-show original form
this.Close(); // and CLOSE this second form...
Related
I tried opening a second form using a button on my main form, but when I close the second window, I can't open it again.
I added the following code to my main form:
settings_window secondForm = new settings_window();
private void settings_button_Click(object sender, EventArgs e)
{
secondForm.Show();
}
But when I try to open the second form named settings_window the second time, I get the following error: System.ObjectDisposedException.
I found the following code to fix this but I don't know where to place it:
private void settings_window_FormClosing(object sender, FormClosingEventArgs e)
{
this.Hide();
e.Cancel = true; // Do not close the form.
}
You can avoid storing references of Forms and use a simple generic method that shows a Form when an instance of it already exists or creates a new one (and shows it) when none has been created before:
private void ShowForm<T>() where T : Form, new()
{
T? f = Application.OpenForms.OfType<T>().SingleOrDefault();
if (f is null) {
f = new T();
f.FormClosing += F_FormClosing;
}
f.Show();
BeginInvoke(new Action(()=> f.WindowState = FormWindowState.Normal));
void F_FormClosing(object? sender, FormClosingEventArgs e)
{
e.Cancel = true;
(sender as Form)?.Hide();
}
}
When you need it, call as ShowForm<SomeFormClass>(), e.g.,
ShowForm<settings_window>()
Note:
This code uses a local method to subscribe to the FormClosing event.
You can use a standard method, in case this syntax is not available.
BeginInvoke() is used to defer the FormWindowState.Normal assignment. This is used only in the case you minimize a Form, then right-click on its icon in the TaskBar and select Close Windows from the Menu. Without deferring this assignment, the minimized Form wouldn't show up again.
When the starting Form closes, all other Forms close as well
This code supposes nullable is enabled (e.g., see object? sender). If nullable is disabled or you're targeting .NET Framework, remove it (e.g., change in object sender)
Is secondForm a private field of the main form class?
It should work then.
Alternative solution is to show it as as modal - ShowDialog()
Also if you want to save some data in your second form, just use some data initialization from constructor, then saving to first/parent form.
I think you need to create a new instance of the form each time you want to open it. It will create a new instance of the settings_window form each time the button is clicked.
private void settings_button_Click(object sender, EventArgs e)
{
// Create a new instance of the form
settings_window secondForm = new settings_window();
secondForm.Show();
}
Your code shows a class that you have named settings_window, which gives us a hint about what its intended use might be. In general, for a form that behaves "like" a settings window, that you can call multiple times using the same instance, you can declare a member variable using this pattern:
public partial class MainForm : Form
{
public MainForm()
{
InitializeComponent();
// Provide some means, like a menu or button, to show 'secondForm'
settingsMenu.Click += onClickSettingsMenu;
// Dispose the settings_form when the MainForm does.
Disposed += (sender, e) => secondForm.Dispose();
}
// By instantiating it here, its public default or persisted
// properties are immediately available, for example even
// while the main form constructs and loads the initial view.
settings_window secondForm = new settings_window();
private void onClickSettingsMenu(object? sender, EventArgs e)
{
if(DialogResult.OK.Equals(secondForm.ShowDialog()))
{
// Apply actions using the properties of secondForm
}
}
}
This is suitable for any form when you want to:
Repeatedly show and hide the form (e.g. a "Settings" form where the user can change the options multiple times).
Retrieve the default or the persisted properties of the form from the outset even if it's never been shown.
Use any of the form's public properties (e.g. GameLevel, SortType etc.) at any given moment while the app is running, even if the form isn't currently visible.
Display the form modally meaning that "no input (keyboard or mouse click) can occur except to objects on the modal form".
The reason it works is that calling ShowDialog (unlike calling Show) intentionally does not dispose the window handle, and this is to support of this very kind of scenario. The app is then responsible for disposing the resource when the app closes.
The Microsoft documentation for ShowDialog explains how this works.
Unlike non-modal forms, the Close method is not called by the .NET Framework when the user clicks the close form button of a dialog box or sets the value of the DialogResult property. Instead the form is hidden and can be shown again without creating a new instance of the dialog box. Because a form displayed as a dialog box is hidden instead of closed, you must call the Dispose method of the form when the form is no longer needed by your application.
I'm working on a c# program and I want a panel to appear on a form when a button is clicked in another. So when the add button is clicked on form2 the panel requesting the details for this to be possible will be displayed on form 1.
I currently have a static method set up in form1 which can be accessed from form2 - however due to panel.Show() being non static it won't allow me to use this in the function.
In Form1 I have:
public static void showPanel()
{
panel.Show()
}
In my second form I have the following:
private void btn_add_Click(object sender, EventArgs e)
{
form1.showPanel();
this.Hide();
}
I have tested with just having the static function show a message box which works. Is it possible to do it the way I want or do I need to take a few steps back and try a different technique?
Are we talking about a new instance of the second form if so you can try to instantiate the new form using:
Form newForm = new YourFormName(potential parameters);
newForm.showPanel();
newForm.Show();
If you want to execute the form on an open form you can give the first form a reference(field with instance) of the second form. Or you can try using: Application.OpenForms. if you give it [1] it'll give you the second open form probably. You can also use .OfType to get the correct form in case your form order isn't always the same.
PROBLEM SOLVED
SHORT STORY
I want to detect "FormClosing()" event through different forms, ie, when form1 is closed that is instantiated within form2, can form2 detect when user presses exit in form1?
LONG STORY
My team and I are working on a windows form application. Project has two forms: one is the main form page and the other is accessed via this main form. Main form looks like this:
And the second one looks like this:
If you press "Ekle/Sil" buttons within the main form, you are directed to form 2 where you can edit database entries. When you press "Sayfayı Yenile" button in the main form, the content of the text areas are refreshed by re-fetching entries from the database.
My problem is, I want to automatically refresh the main form when the user closes the second form. My research suggests I should use an "FormClosing()" event to detect a closing form. However, I want to detect this from the main form. Instantiating main form in second form's source code doesn't seem to be a reliable solution. Anyone can tell me how to do this?
EDIT
I solved the problem:
1) Created a public method within the main form that refreshes the page.
2) Send "this" property from the main form when creating the second form.
3) Added an "FormClosed()" handler within the second form that invokes this public method.
Still, I'm looking for a better solution.
EDIT 2
Better solution InBetween's answer
Simply use the Form.Closed event of the new child windows form. Everything is handled from the main form:
void EkleSil_Clicked(object sender, EventArgs e) //or whatever method is called when button is clicked
{
var newChildForm = new ChildForm();
newChildForm.Closed += childFormClosed;
newChildForm.Show();
}
void childFormClosed(object sender, EventArgs e)
{
((Form)sender).Closed -=childFormClosed;
updateShownData();
}
You can create a event in the second form and raise it when the form is closing .
Handle the event in the main form and refresh the main form when the event is raised
Another option would be to pass Form1 as an argument to Form2. Then use the Form.Closing event in Form2 and use the Form1 reference to trigger something.
Form1 form1Ref;
public Form2(Form1 mainform)
{
form1Ref = mainform;
}
private void Form2_FormClosing(object sender, FormClosingEventArgs e)
{
form1Ref.SomeMethod();
}
I have a form1, from this form1 going to new form2, form2 new form 3... and when I close form1 then application close.but, i want close all form before close application.
you may try
foreach(Form f in Application.OpenForms)
{
f.Close();
}
you may adjust it as needed.
see more on Application.OpenForms
If I understand the question correctly (and I might not...it's hard to read), what you want is to ensure that closing your first form doesn't close the entire program, and that closing the last form does. If that's the case, then you need to make the following changes:
In your Program.cs file, you'll have some code that looks like Application.Run(new Form1());. This passes a Form instance to the Run() method, and the Run() method will automatically return and allow the program to exit when that Form instance is closed.
Instead, don't pass anything to Run(). Use the parameterless overload: Application.Run(); You will still have to create the first Form instance and show it. E.g. before the call to Application.Run(); add this:
new Form1().Show();
Having done that, you now need to call Application.Exit() when you actually want the program to end. So in your last form that you show, override OnFormClosed():
protected override voide OnFormClosed(FormClosedEventArgs e)
{
base.OnFormClosed(e);
Application.Exit();
}
If that's not what you wanted to do, please try to word your question more carefully and explain in more detail what you want.
just write this to show every form
Form1 f=new Form1();
f.ShowDialog();
Class System.Windows.Forms.Form has an event FormClosing. You can handle the closing of forms there.
In form1:
public void form1_FormClosing(object sender, FormClosingEventArgs e) {
form2.Close()
// if we have the instances of form3, form4 here, then:
// form3.Close()
// form4.Close()
}
In form2:
public void form2_FormClosing(object sender, FormClosingEventArgs e) {
form3.Close()
}
etc.
I have problem with Windows Forms.
When I show form2 from form1 like this (names of the variables changed):
form2.ShowDialog(form1);
then I have this exception:
System.InvalidOperationException: Form that is already visible cannot be displayed as a modal dialog
box. Set the form's visible property to false before calling showDialog.
Telling more - during debugging I see that after calling ShowDialog method debugger goes again to the same ShowDialog method - and this is why I have this exception. I suppose that form1 is loaded again and it is some kind of ShowDialog method bug? I have form2 Visible property set to false. I tried to use Hide method too - not working.
Edit:
More info - I use ShowDialog method after showing combo box selector from form1. When I click the last property in combo box by mouse - ShowDialog is working. If I go down by keyboard and click 'Enter' mentioned exception appears.
Try something like this. When you close Form2, control will go back to Form1 (or whoever called Form2):
private void button1_Click(object sender, EventArgs e)
{
Form2 form2 = new Form2();
form2.ShowDialog();
}
Or just
private void button1_Click(object sender, EventArgs e)
{
new form2().ShowDialog();
}