如何使用公共get set从表单2启用表单1中的按钮？

Question

表格1中有两个按钮,一个是"ShowForm2"按钮,另一个是"button1"按钮.

默认情况下禁用按钮1.当我点击"ShowForm2"按钮时,表格2将显示.

所以,我想要的是,当我单击表单2中的"button2"时,它将启用表单1中的"button1".

所以,我尝试在我的form2类中这样编码:

public partial class Form2 : Form
{
    bool enable_form1_button1;
    public bool Enable_form1_button1
    {
        get { return this.enable_form1_button1; }
        set { this.enable_form1_button1 = value; }
    }
    public Form2()
    {
        InitializeComponent();
    }

    private void button2_Click(object sender, EventArgs e)
    {
        enable_form1_button1 = true;
    }
}

然后在我的Form1类中,我希望将"enable_form1_button1 = true"传递给表单1并启用表单1 button1.但是怎么做呢？

public partial class Form1 : Form
{
    public Form1()
    {
        InitializeComponent();
    }

    private void btb_Showfrm2_Click(object sender, EventArgs e)
    {
        Form2 frm2 = new Form2();
        frm2.Show();
        button1.Enabled = frm2.Enable_form1_button1; // I put it here, and it just does not seems right
    }
}

Answer 1

那么你可以做的是,将按钮公开为属性,并将当前表单的引用发送到需要控制的表单:

Form1中

public partial class Form1 : Form
{        
    public Button BtnShowForm2
    {
        get { return btnShowForm2; }
        set { btnShowForm2 = value; }
    }

    private void btnShowForm2_Click(object sender, EventArgs e)
    {
        // pass the current form to form2
        Form2 form = new Form2(this);
        form.Show();
    }
}

然后在Form2中:

public partial class Form2 : Form
{
    private readonly Form1 _form1;

    public Form2(Form1 form1)
    {
        _form1 = form1;
        InitializeComponent();
    }

    private void btnEnabler_Click(object sender, EventArgs e)
    {
        // access the exposed property here <-- in this case disable it
        _form1.BtnShowForm2.Enabled = false;
    }
}

我希望这就是你要找的东西.