maxudaskin Posted November 1, 2007 Share Posted November 1, 2007 <div class="menu_button_l" onmouseover="this.style.background='url(buttonhv.gif)'" onmouseout="this.style.background='url(buttonhv.gif)'">Home</div> It does not do the onmouseout... any ideas? Link to comment https://forums.phpfreaks.com/topic/75598-swap-style/ Share on other sites More sharing options...
cooldude832 Posted November 1, 2007 Share Posted November 1, 2007 Ask your self what the name of this forum is, ask yourself what language that question is, and then realize no one here will help you... Link to comment https://forums.phpfreaks.com/topic/75598-swap-style/#findComment-382530 Share on other sites More sharing options...
kratsg Posted November 1, 2007 Share Posted November 1, 2007 Well, this is easy javascript (and pshaw, JavaScript and PHP go together like.. AJAX :-o) What you have are incorrect CSS javascript references. <div class="menu_button_l" onmouseover="this.style.background='url(buttonhv.gif)'" onmouseout="this.style.background='url(buttonhv.gif)'">Home</div> http://codepunk.hardwar.org.uk/css2js.htm You need to use backgroundImage instead (capitalization counts!) <div class="menu_button_l" onmouseover="this.style.backgroundImage.src='buttonhv.gif';" onmouseout="this.style.backgroundImage.src='buttonhv.gif';">Home</div> Also, it needs to be different images... seriously o_o Link to comment https://forums.phpfreaks.com/topic/75598-swap-style/#findComment-382546 Share on other sites More sharing options...
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