最新回答 / qq_呵呵_9
for (var i = 0; i < tits.length; i++) {tits[i].onmouseover=tab(i);};这个地方不直接用使用方法名称,使用下面的就可以了,tits[i].onmouseover=function tab(i){ alert(i);};
2015-04-13
最新回答 / 慕设计9479289
https://www.imdb.com/list/ls501260222/https://www.imdb.com/list/ls501260222/?mode=desktophttps://www.imdb.com/list/ls501260226/https://www.imdb.com/list/ls501260226/?mode=desktophttps://www.imdb.com/list/ls501260246/https://www.imdb.com/list/ls501260246/?...
2015-04-07