Opening new form is reseting the design? [closed] - c#

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So I have a project with two forms: the MainForm and AnotherForm.
In the MainForm I have a button which should open the AnotherForm. I'm using the following code to do that:
private void btnTimeReminder_Click(object sender, EventArgs e) {
AnotherForm anotherForm = new AnotherForm();
anotherForm.Show();
this.Hide();
}
The problem is, when I click the button to open the "AnotherForm", it just opens a blank form in the WindowsDefaultPosition. All the design of the AnotherForm is deleted (The buttons, images, even the size of the AnotherForm).
I have no ideia why is this happening... Am I using a bad code to open the form?

Ensure that
InitializeComponent();
is being called in the constructor of your new form.
If you need to change the initial position of your new form, you can use
anotherForm.StartPosition = FormStartPosition.CenterParent;
or whichever start position is necessary.

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How to access design Windows Forms C# [closed]

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Pretty much new to Windows Forms, I know the C# language, just not in the same context. I have searched around for a while and it seems to me that every solution is doing something similar to this:
Label1.Text = "I'm a label".
But I don't understand where Label1 is coming from.
All I have is a new Windows Form Application, which comes with one form preloaded and a Program class. So as this class came with some code, I thought this would be a logical way of accessing the label's properties:
static class Program
{
static void Main(String[] args)
{
FormUpdate frmUpdate = new FormUpdate();
Application.EnableVisualStyles();
Application.SetCompatibleTextRenderingDefault(false);
Application.Run(frmUpdate);
// Why isn't this a suitable way of getting the label?
frmUpdate.label1.Text = "I cause an error!";
}
}
But I don't understand where Label1 is coming from.
Someone used the Visual Studio Designer for Windows Forms and dragged and dropped a Label component onto their form. As Visual Studio has no way of naming them, but needs a name, it simply counts up. The first dropped label is called "Label1".
The access specifier for those controls added is private by default and I'd suggest to leave it that way. If you want to interact with your form, either do it from inside your form or write a public method that you call that will then set all the private properties like the text of a certain label.
Generally speaking, Application.Run(frmUpdate); is running the program, based on the starting form you gave. Anything after that will have little effect. So you ran your form and after you closed it, you set the label. That's not going to have any visible effect. You need to do that before you run the form or while you are running it.

The textbox1.Text can't copy the textbox2.Text value from another form [closed]

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Below is my code
private void btnCptKb5_Click(object sender, EventArgs e)
{
Form1 form1 = new Form1();
textbox1.text = form1.textbox1.text;
}
Why is that I can't copy the value of form1 textbox?
You didn't change Textbox2.Text value
You are creating a completely new form instance,and trying to access that form's textbox value.It will be always empty even if your code works.
I think This question is most commonly asked question in StackOverflow You can take a look at these questions, and I'm sure you will find an appropriate answer:
1) How to pass values between forms in c# windows application?
2) Passing data between forms
3) switching between forms without loss of information
4) windows.form c# moving between forms
5) Communicate between two windows forms in C#
6) How to share data between forms?
7) Passing Data Between Forms
8) Get data from one textbox on form1 from another textbox on form2

I want to move from one form to another old already form programmatically [closed]

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I have created two forms in a Windows Forms application using C#. I just want to move from one form to the other. I don't want to open a new form.
I created a LinkLabel, an OnClickEventHandler and in that handler I want to move to already created form programmatically.
Is there any property of LinkLabel so I can set name or address of other form? So that on click it can automatically take me there?
I suppose that you open the second form in a similar way
Form2 f = new Form2();
f.Show();
If this is the case, then you could save the reference to the second form in a global class level variable and use that reference when you need to show again the second form
private Form2 theSecondForm = null;
....
// Open the second form...
theSecondForm = new Form2();
theSecondForm.Show();
When you need to switch to the second form
// Check if the second form is still available
if(theSecondForm != null && !theSecondForm.IsDisposed)
theSecondForm.Show();
else
{
theSecondForm = new Form2();
theSecondForm.Show();
}
Notice, that before calling the second form show method is better to check if the variable is still valid and point to a real instance of the second form. You could also wire to Form Close event to be notified if the user closes the second form
theSecondForm = new Form2();
theSecondForm.Show();
theSecondForm.FormClosed += ClosingSecondForm;
private void ClosingSecondForm(object sender, FormClosedEventArgs e)
{
// Set the global variable to null... this prevent to call a closed form
theSecondForm = null;
}

Cannot see the openFileDialog and saveFileDialog in C# [closed]

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Questions concerning problems with code you've written must describe the specific problem — and include valid code to reproduce it — in the question itself. See SSCCE.org for guidance.
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I am creating a windows form with the openFileDialog and the showFileDialog icons. But when I run the form, I do not see the options in the top left corner.
Is there some properties that I need to change to visible or something?
Any help will be appreciated.
These Dialog are not shown unless you perform any action (either clicking a save button or open file Button)
what you need to do create a button and then handle its click event
something like this.
this.button1.Click += new System.EventHandler(this.button1_Click);
and then in this event, you can invoke your dialog
private void button1_Click(object sender, EventArgs e)
{
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
string filename = openFileDialog1.FileName; // this is the selected file
}
}

Resetting a Windows Forms form's elements to an initialized state (C#/.NET) [closed]

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I'm building a Windows Forms form in C# with various elements in a panel that starts out either invisible, disabled, or set to null (labels, combo boxes, grids, etc.). As the user goes through and makes choices, these elements are populated, selected, etc.
The idea is to upload files, read them, and process entries to a database. Once the processing for this directory has completed, I'd like to be able to have the user select another directory without exiting and restarting the Windows Forms application, by pressing a button that becomes visible when the process has completed.
Is there an easy call to reset the application (or the panel that contains the elements), similar to when a webform is refreshed, or do I have to write a function that "resets" all of those elements one at a time?
As the result of a development meeting, my project has changed direction. I thank the two of you who helped with answers, and am going to close the question.
Simply remove the panel from the form and create the new one.
Sample:
Panel CreatePanelWithDynamicControls() {
Panel ret = new Panel();
ret.Dock = DockStyle.Fill;
// Some logic, which initializes content of panel
return ret;
}
void InitializeDynamicControls() {
this.Controls.Clear();
Panel pnl = this.CreatePanelWithDynamiControls();
this.Controls.Add(pnl);
}
void Form1_Load(object sender, EventArgs e) {
if (!this.DesignMode) {
this.InitializeDynamicControls();
}
}
// I don't know exactly, on which situation
// do you want reset controls
void SomeEvent(object sender, EventArgs e) {
this.InitializeDynamicControls();
}
You could try calling this.InitializeComponent(), which may do the trick. Alternately, if your application has a 'Directory Select' form and a 'Process Files' form, you could have the Directory Select form do a "new" on the Process Files form, which should return it to its original state (not while its open, though).

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