Solved

Let a click "fall through"

Posted on 2007-12-02
10
958 Views
Last Modified: 2012-06-21
I have a PictureBox (MAIN), upon which I am drawing a bunch of smaller PictureBoxes (and assigning as their parent, the larger PictureBox), such that they completely overlap the parent PictureBox.

The problem I have run into now is I have signed-up for the OnClick for the MAIN PictureBox -- but it is not being processed because the click is happening on one of the child PictureBoxes and is never "reaching" the Parent (MAIN) PictureBox.  How can I let the click "fall through" and get processed by the Parent?
0
Comment
Question by:Tom Knowlton
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 5
  • 4
10 Comments
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20391810
on the child's picturebox event, you can retrasmit the event manually. Derive your own picturebox class




public class ChildPictureBox : PictureBox
{
      protected override void OnClick(EventArgs e)
      {
            if (this.Parent != null)
                 Parent.OnClick(e);
            base.OnClick(e);
      }
}

Open in new window

0
 
LVL 5

Author Comment

by:Tom Knowlton
ID: 20392826
And what coordinates will be passed through to the Parent?  Will it be the relative coordinates of the child, or the coordinates of the Parent?
0
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20392931
hmm, them will be relative to the child picture box, but you can convert with an ugly trick:

notice the onclick method doesn't pass the mouse cursor coordinates. Use the OnMouseClick instead, with some conversion tricks:
 
protected virtual void OnMouseClick (MouseEventArgs e)
{
     base.OnMouseClick(e);
     if (this.Parent != null)
     {
           e.Location = Parent.PointToClient(this.PointToScreen(e.Location));
           Parent.MouseOnClick(e);
     }
}

Open in new window

0
Technology Partners: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 5

Author Comment

by:Tom Knowlton
ID: 20451509
can you please paste in code that shows how this works with a derived picturebox?
0
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20452088
create your derived class as shown above. The control will appear on the toolbox when designing your form, use it instead of PictureBox.
In your form, handle the Click event, it will be fired every time the picture box is clicked.


public class MappedPictureBox : PictureBox
{
    protected virtual void OnMouseClick (MouseEventArgs e)
    {
       base.OnMouseClick(e);
       if (this.Parent != null)
       {
           e.Location = Parent.PointToClient(this.PointToScreen(e.Location));
           Parent.MouseOnClick(e);
       }
    }
}

Open in new window

0
 
LVL 5

Author Comment

by:Tom Knowlton
ID: 20452185
which kind of "new item" do I pick from "Add New Item"?

That is part of my confusion is - how / where do I create this derived control such that it will appear on the Toolbox?
0
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20452258
if you create the class and include it in your project, just like other .cs file. The control will automatically appear in the toolbox.
0
 
LVL 5

Author Comment

by:Tom Knowlton
ID: 20452304
I still don't see it in the Toolbox, sorry.
0
 
LVL 55

Accepted Solution

by:
Jaime Olivares earned 500 total points
ID: 20454371
Hmm, there are many things to consider. First you have to compile to allow the control appear in your toolbox. But the code won't compile as is, you will need to derive  the main picture box to allow the code to work, but you can reuse it this way:

namespace System.Windows.Forms
{
    public class MappedPictureBox : PictureBox
    {
        protected override void OnMouseClick(MouseEventArgs e)
        {
            base.OnMouseClick(e);
            if (this.Parent != null && this.Parent is MappedPictureBox)
            {
                Point p = Parent.PointToClient(this.PointToScreen(e.Location));
                MouseEventArgs mea = new MouseEventArgs(e.Button, e.Clicks, p.X, p.Y, e.Delta);
                ((MappedPictureBox)Parent).OnMouseClick(mea);  // perform the click
            }
        }
    }
}

Open in new window

0

Featured Post

What does it mean to be "Always On"?

Is your cloud always on? With an Always On cloud you won't have to worry about downtime for maintenance or software application code updates, ensuring that your bottom line isn't affected.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Entity Framework is a powerful tool to help you interact with the DataBase but still doesn't help much when we have a Stored Procedure that returns more than one resultset. The solution takes some of out-of-the-box thinking; read on!
Exception Handling is in the core of any application that is able to dignify its name. In this article, I'll guide you through the process of writing a DRY (Don't Repeat Yourself) Exception Handling mechanism, using Aspect Oriented Programming.
There are cases when e.g. an IT administrator wants to have full access and view into selected mailboxes on Exchange server, directly from his own email account in Outlook or Outlook Web Access. This proves useful when for example administrator want…
Visualize your data even better in Access queries. Given a date and a value, this lesson shows how to compare that value with the previous value, calculate the difference, and display a circle if the value is the same, an up triangle if it increased…
Suggested Courses

632 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question