Windows forms перейти на другую форму

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12 years, 5 months ago

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I’m using winforms now.
I have the main form «form1» and I have a button that opens form2.
When I open form2 I would like for form1 to disappear. When the user click the x button on form2 I would like for it to close and go back to form1. I wouldn’t like to use modal windows.

winforms

Corey Adler

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asked May 1, 2011 at 11:47

Itay.BItay.B

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2 Answers

    private void button1_Click(object sender, EventArgs e) {
        var frm = new Form2();
        frm.Location = this.Location;
        frm.StartPosition = FormStartPosition.Manual;
        frm.FormClosing += delegate { this.Show(); };
        frm.Show();
        this.Hide();
    }

answered May 1, 2011 at 11:56

Hans PassantHans Passant

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Nice one, I was using owner property up till this point but this would do it and actually much easier.

Jan 30, 2012 at 1:17

In order not to change Form’s Properties, simply use the ShowDialog() method in Form1.cs code to open Form2. That will deactivate Form1:

void OpenSecondForm()
{
  Form form2 = new Form();
  form2.ShowDialog();
}

KernelPanic

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answered Nov 18, 2016 at 12:09

I think there is a typo. sb Form form2 = new form2();

Aug 5, 2021 at 23:58
Pat is right, so anyone using this example should bear that in mind.

Jun 19 at 14:41

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2 Answers

    private void button1_Click(object sender, EventArgs e) {
        var frm = new Form2();
        frm.Location = this.Location;
        frm.StartPosition = FormStartPosition.Manual;
        frm.FormClosing += delegate { this.Show(); };
        frm.Show();
        this.Hide();
    }

answered May 1, 2011 at 11:56

Hans PassantHans Passant

924k146 gold badges1696 silver badges2536 bronze badges

Nice one, I was using owner property up till this point but this would do it and actually much easier.

Jan 30, 2012 at 1:17

In order not to change Form’s Properties, simply use the ShowDialog() method in Form1.cs code to open Form2. That will deactivate Form1:

void OpenSecondForm()
{
  Form form2 = new Form();
  form2.ShowDialog();
}

KernelPanic

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answered Nov 18, 2016 at 12:09

I think there is a typo. sb Form form2 = new form2();

Aug 5, 2021 at 23:58
Pat is right, so anyone using this example should bear that in mind.

Jun 19 at 14:41

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Переходы между формами

Вопрос
Всем привет. Возникла такая проблема.

Есть две формы, необходимо сделать так, чтобы по клику на кнопку на первой форме — появлялась вторая, а первая скрывалась. А по клику на кнопку на второй форме — вторая закрывалась, а первая появлялась.

Пробовал делать несколькими вариантами:

Первый вариант:
```
//кнопка на первой форме 
private void button1_Click(object sender, EventArgs e)
        {
            Form f3 = new Form3();
           
             f3.Show();
           button1.Enabled = false;
           this.Visible = false;
            
            
        }
//евент, когда форма становится активной, чтоб визибл был тру
  private void Form2_Activated(object sender, EventArgs e)
        {
            this.Visible = true;
        }

//кнопка на второй форме
 private void button2_Click(object sender, EventArgs e)
        {
            Form2 f2 = new Form2();
            f2.Activate();
            this.Close();
        }
```
Данный вариант не работает — первая форма не выходит из состояния невидимости.

Вот второй вариант:
```
// кнопка на первой форме 
Form f3 = new Form3();
           
             f3.Show();
           button1.Enabled = false;
           this.Hide();
//кнопка на второй форме
  private void button2_Click(object sender, EventArgs e)
        {
            Form2 f2 = new Form2();
            f2.Show();
            this.Close();
        }
```
Этот вариант работает, но неверно. Почему-то на команду
```
 f2.Show();
```
Создаётся новый экземпляр Form2. А Form2 — главная форма приложения и в данном случае закрывая заново созданную форму 2 — приложение не закрывается, т.е. изначально созданная Form2 висит в невидимости.

Извиняюсь за несколько путанное объяснение=) Помогите пожалуйста исправить ошибку.

Ответы

Вот рабочий вариант.
```
//Form1.cs
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace WindowsFormsApplication1
{
  public partial class Form1 : Form
  {
    private Form2 form2;
    public Form1()
    {
      InitializeComponent();
      form2 = new Form2(this) { Visible = false };

    }

    private void button1_Click(object sender, EventArgs e)
    {
      this.Visible = false;
      form2.Visible = true;
    }
  }
}
```
```
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace WindowsFormsApplication1
{
  public partial class Form2 : Form
  {
    private Form1 _form1;
    public Form2(Form1 form1)
    {
      _form1 = form1;
      InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {
      this.Visible = false;
      _form1.Visible = true;
    }
  }
}
```
Т.е. в проекте создаёте две формы, причём первая будет главной, а вторая второстепенной. При инициализации первой формы(т.е при запуске приложения) создаёте экземпляр второй формы, скрываете её и добавляете туда ссылку на текущую(т.е) первую форму. А потом
при клике скрываете первую показываете вторую, показываете первую скрываете вторую в другом обработчике(второй формы).
- Помечено в качестве ответа
  
  10 марта 2012 г. 5:28

Источник

I have some code here:

private void button1_Click(object sender, EventArgs e)
{
    Application.Run(new Form3());
}

Although I don’t think this is how you are meant to change forms, when I ran it, it threw an error stating:

Starting a second message loop on a
single thread is not a valid operation

H H

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asked Jun 15, 2011 at 21:14

You cannot use Application.Run — that is for starting windows form application (internal message loop which is shared among all forms in the application), not for showing a form. Each form has Show and Hide method so you should simply call:

private void button1_Click(object sender, EventArgs e)
{
    Form3 f = new Form3(); // This is bad
    f.Show();
}

But you should not create form each time you want to show it. If you want to have only one instance of the form you should keep it as global and only show or hide it on demand. Otherwise you will have to call Close instead of Hide to clear all resources the form consumes.

answered Jun 15, 2011 at 21:20

Ladislav MrnkaLadislav Mrnka

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you can do as simple as it is :

test mp = new test();
mp.Text = " Welcome Mr." + textBox1.Text;
this.Hide();
mp.ShowDialog();
this.Close();

where test is your new form
and then if you wona pass variables to the other form just make you variable as public then you can make :

 mp.Text = " Welcome Mr." + textBox1.Text;

then showdialog to show your new form and close the old form..

natiiix

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answered Jun 15, 2011 at 21:21

TamerTamer

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Your first form should create the second form on the first usage and hide itself from view. Then the second form will hide itself and show the first form when you switch back. Something like this…

  private Form _firstForm;
  private Form _secondForm;

  private void switchToSecond_Click(object sender, EventArgs e)
  {    
      if (_secondForm == null)
          _secondForm = new SecondForm();

      Hide();
      _secondForm.Show();
  }

  private void switchToFirst_Click(object sender, EventArgs e)
  {    
      Hide();
      _firstForm.Show();
  }

You would assign to firstForm the this value inside the constructor of the first form class and because this is the main form it will always be created at application startup. If you want to save resources you could also dispose of the second form when you switch back to the first and so recreate it each time you switch.

answered Jun 16, 2011 at 5:19

Phil WrightPhil Wright

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this.Hide(); // Hide the current form.

Menu menu = new Menu(); // Create new instance of the new form.

menu.Show(); // Show it

Andrew

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answered Jan 9, 2018 at 0:52

Assume you have Form1 and Form2
Your scope is not clear if you want

Switch between Form1 and Form2 and keep both of them alive.
Keep only either Form1 or Form2 at a time.

for case 1 simple Form1.Hide() and Form2.Show() would be adequate.

for case 2 the solution will be a little bit complicated.

Create your own instance of ApplicationContext MyAppCxt= new ApplicationContext().
instead of the default Program.Run(Form1) call Program.Run(MyAppCxt(Form1))
in the handler of some event, when you need to switch the Form1 and Form2

Form2 f2 = new Form2();//Create the new form MyAppCxt.MainForm = f2;//set Form2 as MainForm in the message queue Form2.Close()//Close Form2 and release all the resource F2.Show();//display Form2 to the user

the same scenario would be repeated when required to switch back between Form2 and Form1

answered Apr 4, 2019 at 10:34

Источник

Windows forms перейти на другую форму

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