適應另一種答案我有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事件我認爲,當來自移動鼠標僅發生某處到某處。如果我理解正確,當鼠標位於id1中時,應該調用代碼中的hideall(),並移出id1外部。你是否聲稱當鼠標從id1(但不是id2) - > id2移動時,hideall()也會被錯誤地觸發?或者當它從id2-> outside_id2移動時? – ninjagecko 2011-04-16 20:34:03
當我從id1> id2和id1> body移動鼠標,然後調用hideall()函數。我需要那個hideall()函數調用只是如果id1>正文,而不是id1> id2 – somedude 2011-04-16 20:43:07
我已經看到一個非常像這樣的問題:http://jsfiddle.net/userdude/gsv7n/1/ – 2011-04-16 21:38:02