2 replies to this topic

[Darkyere]

I have made this application where theres a tlistview whit image destinations in it.

Then i got a timer to shift the picture.

then ive addet a random function to shuffle through the picture.

now what i want to happen is that the same picture dont get shown twice.

So what i tried was to put the number of the picture in another tlistview and make a "While do" to check if randomly generated number is shown in the list.

this is the code ive tried using.

		    randomnr := random(listview1.Items.Count -1);
                    
		    x := 0;

                    While x<listview2.Items.Count - 1 do
                      begin

                        if listview2.Items[x].Caption = inttostr(randomnr) then
                          begin

                            randomnr := random(listview1.Items.Count -1);

                            x := 0;

                          end;

                        x := x +1;

                      end;

            item := listview1.Items[randomnr];
            form2.image1.Picture.LoadFromFile(item.Caption);

            listview2.Items.Add.Caption :=  inttostr(randomnr);

This is how the form looks like
img.photobucket.com /albums/ v291/ Rambus666/ Programmering/ MultipleFilesDialog.jpg

All help is apreciatet :)

Best regards,
Darkyere

Edited by WingedPanther, 19 November 2008 - 01:17 PM.
add code tags (the # button)

[WingedPanther]

What I would do is have a second listview that is hidden and shift the items from the one you are searching on to the hidden one. That way, you cannot get the same image twice.

[jakefrog]

You just check if the actual number is in a list or array that you can store a number already found!