這裏是我的完整代碼,當我將鼠標懸停在popupcontact div上時,它顯示了divtoshow div,它有一個鏈接名稱rahul,當我將鼠標懸停它隱藏div名稱divt的鏈接。 我的div應該隱藏,當我mouseout不當我滑過鏈接。 請儘快幫助。我需要停止DIV消失時,我將鼠標移到DIV鏈接
問候 拉胡爾
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" dir="ltr">
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.0/jquery.min.js" type="text/javascript"></script>
</head>
<body>
<div id="popupContact" style="position:absolute;left:100px;top:100px;width:100px;height:50px;background-color:orange;border:1px solid red ;">
</div>
<div id="divtoshow" style="display:none;background-color:green; border:1px solid black;width:200px;height:100px;position:absolute;">
dsfdssd <div><a href="#">rahul</a></div>
</div>
</body>
</html>
<script type='text/javascript'>
$(document).ready(function(){
var popup_pos=$('#popupContact').offset();
var timer;
$("#popupContact").mouseover(function() {
if(timer) {
clearTimeout(timer);
timer = null
}
timer = setTimeout(function() {
console.log($("#VersionSelectField").is(':hidden'));
if(!$("#VersionSelectField").is(':hidden')){
$("#divtoshow").css('position',"absolute");
$("#divtoshow").css('top',popup_pos.top-20);
$("#divtoshow").css('left',popup_pos.left-20);
$("#divtoshow").fadeIn(300);
$("#popupContact").hide();
}
}, 100);
});
$("#divtoshow").mouseout(function() {
if(timer) {
clearTimeout(timer);
timer = null
}
timer = setTimeout(function() {
$("#divtoshow").fadeOut("slow");
$("#popupContact").show();
}, 1000);
});
});
</script>
嗨查克,你編輯? – XMen 2010-09-07 06:00:49