CODEWITHSUNDEEP
c#
crazymonster
crazymonster

Reputation: 1

Visible Property of Picture box not working from another form in C#

Hi I am using windows forms in C#. I am trying to modify the visible property of a picture from main form to another. Initially, the visible property of the picture box is set to false. On a button click from another form, the visible property of the picture box is modified to true.

This is the code written in the Form2 method:

private void button_Click(object sender, EventArgs e)
{
    public Form1 frm1 = new Form1();
    frm1.pictureBox.Visible= true;
}

Upvotes: 0

Views: 1606

Answers (3)

serene
serene

Reputation: 685

According to your written code it will create new instance of the desired form, and NOT take the existing open form. Hence to identify the existing open form containing target picture box you need the target form and controlling form be related by like Parent form or MDI Parent Form. Assuming case of MDI Parent Form (i.e. Controlling form is MDI Parent of Target Form), you need following codes to identify to existing open form:

foreach (Form frm in MdiChildren)
        {
            if (frm is myTargetForm)
            {
                //do your code to find control using id of picture box and change the required properties
            }

        }

Upvotes: 0

Sam I am says Reinstate Monica
Sam I am says Reinstate Monica

Reputation: 31206

Form1 is an instance type, so when you do 

public Form1 frm1 = new Form1();
frm1.pictureBox.Visible= true;

you're really just creating a new instance of Form1 completely unrelated from your original Form1, changing a picture-box's visible property on it, and then discarding it.

What you can do, is put a reference to the "parent" Form1 inside your Form2 class.

Here's an example

public partial class Form2 : Form
{
    public Form2(Form1 parent)
    {
        InitializeComponent();
        this.Parent = parent;
    }

    Form1 Parent;

    private void button1_Click(object sender, EventArgs e)
    {
        Parent.pictureBox.Visible= true;
    }
    ...
}

Upvotes: 2

Laurent S.
Laurent S.

Reputation: 6947

there you create an instance of a form : 

public Form1 frm1 = new Form1();

This is then obviously NOT the form you already may have in your page, which you could simply access by its ID.

Upvotes: 0

Related Questions