适应另一种答案我有previously:
<div id="id1" onmouseout="showTarget('id1',0);" onmouseover="showTarget('id1',1);" style="border:1px solid red; width: 150px; height: 100px">
<div id="id2" onmouseover="showTarget('id1',1);" onmouseout="showTarget('id1',1);" style="background-color: red; color: white; width: 100px; height: 75px;">This is inside id1 div</div>
</div>
function showTarget(target, state) {
switch (state) {
case 1:
state = 'block';
break;
default:
state = 'none';
}
console.log(state);
document.getElementById(target).style.display = state;
}
http://jsfiddle.net/gMpkX/
mouseout事件我认为,当来自移动鼠标仅发生某处到某处。如果我理解正确,当鼠标位于id1中时,应该调用代码中的hideall(),并移出id1外部。你是否声称当鼠标从id1(但不是id2) - > id2移动时,hideall()也会被错误地触发?或者当它从id2-> outside_id2移动时? – ninjagecko 2011-04-16 20:34:03
当我从id1> id2和id1> body移动鼠标,然后调用hideall()函数。我需要那个hideall()函数调用只是如果id1>正文,而不是id1> id2 – somedude 2011-04-16 20:43:07
我已经看到一个非常像这样的问题:http://jsfiddle.net/userdude/gsv7n/1/ – 2011-04-16 21:38:02