在 C# 中，当我单击第二个表单的按钮时，如何在第一个表单中打开第三个表单

Question

In my Windows Forms application, I have 3 different Form s ( Form1, Form2, Form3 )

• Form1 Contains a Button and a Panel (button1, panel1)

  

• Form2 Contains only a button (button)

  

• Form3 Contains nothing

  

In Form1 when I click button1 , Form2 opens in panel1

private void button1_Click(object sender, EventArgs e)
{
    panel1.Controls.Clear();
    Form2 f2 = new Form2();
    f2.TopLevel = false;
    panel1.Controls.Add(f2);
    f2.Dock = DockStyle.Fill;
    f2.Show();
}

Now, inside Form2, when i click the button, I want the Form3 open inside Form1's panel, I've tried this code...

Form1 f1 = new Form1();
private void button1_Click(object sender, EventArgs e)
{
    f1.panel1.Controls.Clear();
    Form3 f3 = new Form3();
    f3.TopLevel = false;
    f1.panel1.Controls.Add(f3);
    f3.Dock = DockStyle.Fill;
    f3.Show();
}

Note: In Form1 I've changed the modifier to public for the Panel

Answer 1

Step 1: Create functions in Form1 to Show the other Forms:

class Form1
{
    public void ShowForm2() {...}
    public void ShowForm3() {...}
    // TODO: think about: do you need IsForm2Shown?
    // TODO: think about what you want if ShowForm2 is called twice
    ...

}

Step 2: Let Form2 know about Form1. Give Form2 a function to show Form3 inside Form1

class Form2
{
    public Form1 Form1 {get; set;}

    protected void ShowForm3InsideForm1()
    {
        this.Form1.ShowForm3();
    }
}

Whenever Form one creates Form2 it should fill property Form1 . So inside Form1.ShowForm2() :

var form2 = new Form2()
{
   Form1 = this,
   ...
}
form2.Show();

Step 3: whenever you find in Form 2 that you want to "show form 3 in form 1", just call the proper function

// Somewhere inside Form2 code you decide that you want to show Form3 inside Form1:
this.ShowForm3InsideForm1();

Simple comme bonjour!