Hi Jimi,
so far this was more guessing than knowing from my side.
However, it sometimes help.
Seems, working with activewindow.width leeds to nothing.
But, I managed to position a toolbar containing 6 standard
icons with it's right edge at the rightmost pixel
on an 1024 pixels wide screen. The width of the toolbar
is at a 1024 pixel screen = 144. Makes 24 pixel
for each icon. The left of a 6 icon toolbar is then at 880.
Astonishingly simple, once you know, isn't it?
So if you want to display 10 icons, left would be 1024 - 10*24.
So if you want to display 1 icon, left would be 1024 - 24,
plus some offset for the x in right hand top corner of the toolbar
plus for the down arrow indicated left ot it, if you want to show it.
But that's the fun bit.
Greetings from Bavaria, Germany
Helmut Weber, MVP
"red.sys" & chr(64) & "t-online.de"
Word XP, Win 98
http://word.mvps.org/
>Hi Helmut,
>
[quoted text clipped - 47 lines]
>application.activewindow.width return only half the true width of the screen
>so that the value must therefore be doubled?)
jimi_hendricks - 09 Jun 2005 11:27 GMT
Dear Helmut
Thanks for your additional help. You're right it's simple once you know how!
I really need to implement a solution that works with different screen
resolutions, so i'll keep hunting for a way to get this.
It seems strange that Application.Width and ActiveWindow.Width don't give
this reliably.
Thanks again,
jimi
> Hi Jimi,
>
[quoted text clipped - 20 lines]
> Word XP, Win 98
> http://word.mvps.org/
jimi_hendricks - 09 Jun 2005 11:40 GMT
Hi Helmut,
I found that System.HorizontalResolution did the trick.
Thanks for your ongoing help.
Regards,
Jimi
> Hi Jimi,
>
[quoted text clipped - 72 lines]
> >application.activewindow.width return only half the true width of the screen
> >so that the value must therefore be doubled?)