I still have the problem and been messing with the code to try and get a fix to it.
I wonder if I can use a if statement on events something like if mouse is over image then appear a menu else if mouse is over menu keep menu showing else if mouse is not over the image or menu then fade out else menu hide. I want something like this. DO you think it's possible to use jquery functions in this way?? On Dec 29 2008, 12:43 pm, Aaron <shyhockey...@aol.com> wrote: > HI, I am making a website kinda like a social networkng site. > > I am trying to make a hover effect when the mouse is over the users > image I want to pop up a image menu like upload image and edit image > etc. > > I so far do have the codes working. The problem I face is... when the > user puts the mouse over his image a menu pops up not into a new > window but like a image is viewable which is supposed to happen. > > So if your mouse dosen't go over the image meaning the new popup menu > then the menu stays at that position until your mouse goes over the > image and back out it would fade out the menu. > > What I want is to have a menu to fade out when the mouse is either > off the image or off the menu so then the menu would fade out. > > So in my code so far I was able to fade in the menu when the mouse is > on the user image. But in order to fade out the menu you will have to > put your mouse on the menu and then make the mouse go off the menu and > it would fade out. If you had your mouse on the user image and slide > the mouse off the user image no fade out effect will happen to the > menu. > > Hope this explains what I want. Which is to have a fade out effect on > 2 events I wounder if I should use a IF statement. > > I want to fade out only if the mouse is not on either the user image > or the menu itself.
