Hi, I use window.open to open a new thin window with a list of images to select one of them. Because the window is thin it can easily be covered by other windows. Is there a way I can make that thin window be always on top. I want to be able to make selections on the original window while the thin image selection window is open.
In the body tag of your pop up you could try this:
Every time the window is hidden it should bring it back to the top.
That would work, but it would get annoying fast.
Thanks, it works but while the pop up is open I can’t make selections on the window that invoked it. It works will if I the user had to use the pop up and close it before using the invoking window which unfortionatly not the case here The user should be able to work on the invoking window while the pop up is still open.
So… you don’t really want the popup to be always on top - you want the popup to be on top when the parent window loses focus (i’m guessing ;)). However, one of the problems is that the parent window receives blur events bubbling up from child elements on the page.
Hi, I want one of two things:
Either the small pop up being always on top while still being able to make selections on other open applications and browser windows. Remember that the pop up is small and thus can’t cover the whole screen which leaves space for other windows and apps to show with it. This will be similar to an MSN messenger text conversation window when selecting “tools” then “always on top”.
Or when ever the invoking browser window has focus the pop up goes on top but still be able to make selections on the invoking window.
“<body onBlur=“window.focus()”>” -> Nice code requestcode
Now is there a way to keep the window always on top, BUT, still be able to ‘work’ in a window below, whether it be an IE window, or Word, Excel, etc… ???