Calling Form.Show() in series - c#

everyone.
How do I call a number of forms one after another (meaning call new form after previous one is closed).
I wrote something like this;
Form1.Show();
Form2.Show();
and it naturally results in opening both forms simultaneously. How to make function to wait for the fist form to be closed?

Open one form, handle the FormClosed event, and open the second form from within that. Note that this is only necessary because your forms are not modal. If you were using ShowDialog() the call would not return until the first form was closed and your code would work as it is currently structured.
If you require non-modal behavior (i.e., your user must be able to interact with the owner form while the owned form is open) then use something like this:
// very simplistic example...
Form1 frm = new Form1();
frm.FormClosed += delegate { new Form2().Show(); }
frm.Show();

Related

Prevent click of Form 1 when Form 2 is open in C# [duplicate]

This question already has answers here:
How to prevent clicks outside the current form?
(2 answers)
Closed 1 year ago.
I am creating an application where I have a Form 1 (Main Form). Inside this main form there is a button that opens up a different and smaller form (Form 2). I don't want to hide Form 1 when showing Form 2 but because of the size of Form 2, the user is able to click Form 1 hiding Form 2 with its size (not hidden like Form2.hide();) however this is something I don't want to do. I want to prohibit the user to click Form 1 if Form 2 is currently open. Is there any event or function I can use to do this? I have seen in it in other applications but I don't even know how to look for it.
In addition to using the ShowDialog method as suggested in question's comments, you can disable the form itself if you want for example the user be able to switch between forms and view or copy some text.
Form.ShowDialog Method: Shows the form as a modal dialog box.
Here is a definition:
Wikipedia: A modal window creates a mode that disables the main window but keeps it visible, with the modal window as a child window in front of it. Users must interact with the modal window before they can return to the parent application. This avoids interrupting the workflow on the main window. Modal windows are sometimes called heavy windows or modal dialogs because they often display a dialog box.
If you don't want to use a modal form, you can initialize the form2 instance like that:
form2.FormClosed += (_s, _e) => this.Enabled = true;
Thus now you can call:
this.Enabled = false;
form2.Show();
You can also check the ShowInTaskBar property of the form2.
Be careful not to add the lambda event handler several times to the same instance: if form2 is just hidden on close, only one FormClosed += is needed, but is required for every instance of a form you want to manage this way.
If you need to disable only certain controls, use them instead of this:
private void SetControlsEnabled(bool state)
{
myControl1.Enabled = state;
myControl2.Enabled = state;
myPanelHavingSomeControls = state; // this changes all inners too
...
}
form2.FormClosed += (_s, _e) => SetControlsEnabled(true);
SetControlsEnabled(false);
form2.Show();
This is the exact use case for a modal form, which disables the parent form and provides visual clues from the OS to tell the user of that relationship. Something along the lines of this will do:
//Code in the event handler that opens the child form
using (Form2 childForm = new Form2())
{
//Put any initialization code here
childForm.ShowDialog(this);
//Any cleanup or using some return value from the form after it closed
}

New form disappear after creation

I'm facing a problem with form show. I have a main form where I have my GUI and I choose an option that creates an instance of a form. For example in my main form I have:
Form2 f2 = new Form2();
f2.Show();
The problem is that the form shows for about 1-2 secs and then goes behind the main form.
I tried some instructions in the main form below f2.Show() command like
f2.BringtoFront();
this.SendtoBack();
Also I added commands to the new form (Form2) load method:
this.BringtoFront();
this.Activate();
this.Focus();
Nothing of the above commands seems to be a solution for this. Only when I use f2.ShowDialog(); instruction in my main form but I don't want to do that because I need immediate access to both of these forms at the same time.
Any help? Thanks
If you don't want your second form to never go behind your main form then pass the owner in the overload of Show method that accepts the owner parameter
Form2 f2 = new Form2();
f2.Show(this); // Assuming this code runs inside the main form
If you remove or change it to comment this.SendtoBack(); :
f2.BringtoFront();
//this.SendtoBack();
it'll be OK!

Better way to Display forms one on top of another

I have a Main form.I want to launch another form from it and launch another from the launched form.I want to ensure that the Main form is not editable when sub forms are displayed so i use showdialog()
Mainform>(Showdialog)>form1>(showDialog+dispose)>form2(dispose)>Mainform
From Mainform i call form2.ShowDialog() then from form2 i use the following code to launch another form
this.visible=false;
form3.showdialog();
this.dispose();
But there are some problems in this.Is there a better way to achieve what im looking for?
edit:more description
I have a Main form,User clicks a Button on Mainform>Form1 is lauched>User clicks a Button in Form1>Form 2 is lauched(diposing/hiding form1) after form2 is closed Mainform should be brought to front and made editable,until then all other forms should be on top of Mainform and Mainform should be un-editable
The problem is that you have to specify the MainForm as the parent for (both) form2 and form3. When you use the overload of ShowDialog that has no parameters, WinForms uses the active form as the parent, so form3's parent becomes form2 automatically. You are then trying to close/dispose form2 causing form3 to become orphaned.
There are several options for getting the reference to MainForm, but the simplest is to use:
form2/3.ShowDialog(Application.OpenForms["MainForm"]);
Assuming that you have set the Name property on MainForm to "MainForm".
In your code, this.dispose() is executed only after form3 is closed. What i think you want is to close form2 after form3 was closed, so you can call this.Close() instead of this.Dispose().
this.visible=false;
form3.showdialog();
this.Close();
Or maybe, after form3 is shown up you dont need form2 any more. That meas:
this.visible=false;
//show instead of showdialog so it wont wait until form3 is closed
form3.show();
this.Close();
It looks like you are trying to implement something like a wizard. The best solution would be to launch all the child forms sequentially, in the main form.
If you need to pass data along the sequence, you should pass it from each dialog to the main form, which then passes it to the next dialog.
MainForm:
Form1 f = new Form1();
if (f.ShowDialog(this) == DialogResult.OK) {
Form2 f2 = new Form2();
f2.ShowDialog(this);
}
Form1 (button click which will open the form 2):
button1_click(object sender, EvengArgs e) {
this.DialogResult = DialogResult.OK;
Close();
}

Using form.ShowDialog() after this.Dispose();

I'm launching a form from another form. I want the parent form to be disposed before showing the subform, so I use:
this.Dispose();
form.ShowDialog();
The form is displayed, is this the correct way to do it? The form from is launched by a call to ShowDialog() from the main form. I want these forms to be closed before the main form is editable.
EDIT:
This is the Basic Process
Mainform>(Showdialog)>form1>(dispose+showDialog)>form2(dispose)>Mainform
If you want to ensure the parent is not visible before the child is shown, AND you want the parent to Close:
var f = new ChildForm();
this.Hide();
f.Show();
this.Close();
ShowDialog() means, "display this form and return to me after it's closed". That is not what you described you want to happen.
Based on your comment, it might make more sense to create and show form2 from your main form.
In the main form:
DialogResult form1Result;
using (var f = new Form1())
{
form1Result = f.ShowDialog();
}
if (form1Result == DialogResult.OK)
{
using (var f2 = new Form2())
{
f2.ShowDialog();
}
}
And in Form1 event where you originally spawned Form2
this.DialogResult = DialogResult.OK;
this.Close();
Calling Dispose on any object (including the current object known as this) is saying you have finished with that object, it is terminally finished with.
Better to Close the first dialogue and then show the second one.
If you are showing the child dialog modally (which you are in this case) then you can hide the parent dialog What you can do is hide the parent form, show the child form, and then close the parent form.
var form = new SomeForm();
this.Hide();
form.ShowDialog(); // the code will be blocked at this point until the child form closes
this.Close();
this.Dispose(); // i'm not sure if this is necessary anymore. check MSDN
If you are showing a form (not modally), you can hide the parent form, and listen to the child form's close event and then close the parent form in the handler
void childFormClosed(Event e)
{
this.Close();
}
var form = new SomeForm();
form.Closed += new EventHandler... // I usually rely on intellisense with this
this.Hide();
form.ShowDialog();
sorry, I think you cant do that, because when you call ShowDialog method, the child form becomes dependent on its parent form. so the call of dispose will be suspended until you finish with the child form.
I think you have to call the "Show" method and after that you can call the dispose method.
The lifecycle of the child form is depends on its parent lifecycle, that means, once you invoke the parent.close of the parent form, it will dispose all objects created by that instance.
You have to create the child independently from the parent.

C# - window form application - Close the application

I want a help. I have a window form application. Whenever I click on the "close" of the form, the application should itself close.
Can anyone help me.
Regards,
Justin Samuel.
After your explanation:
In Form1, do something like:
Form2 f2 = new Form2();
f2.ShowDialog();
this.Close();
Application.Exit();
This is assuming Form2 can be shown Modal (Dialog), which I think is correct
A Winforms application will exit automatically when the main form is closed. The main form is the form that is instantiated and passed to Application.Run method in the Main method of the application.
If the application process does not exit when this form is closed, something is preventing it from closing, such as a thread (that is not a background thread) that is performing some work.
By your description, it sounds like you may have multiple forms in your application and one form is still running even if its's not visible. In this case, the close button, would close the form, but not exit the application.
You need to add an event handler for the Form_FormClosed event and then call Application.Exit() - Ideally, you could also handle the Form_FormClosing event if you want to give the user a dialog to ask if they really meant to close.
Maybe you are looking for the Application.Exit() method.

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