我有3个id为1,2,3的div,我想要的是在id为1的div上显示id为2的div,当div 2上的鼠标输出事件时显示div 1,以及该div如果我们鼠标离开div 3再次显示div 1并且我们再次鼠标悬停div 1 div 3,则在id为btn n的div 2上单击按钮时出现3。他们必须出现在同一个地方...使用javascript plz ...这里是简单的HTML代码...
<div id="1" style="position:absolute; height:200px; width:200px;">
<table width="100%" border="01">
<tr>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td> </td>
<td> </td>
</tr>
</table>
</div>
<div id="2" style="position:absolute; height:200px; width:200px;">
<table width="100%" border="01">
<tr>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td><input type="button" id="btn"></td>
<td> </td>
<td> </td>
</tr>
</table>
</div>
<div id="3" style="position:absolute; height:200px; width:200px;">
<table width="100%" border="01">
<tr>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
</tr>
</table>
</div>
答案 0 :(得分:0)
尝试
$( "#div2" ).mouseover(function() {
$("#div1").hide();
$("#div2").show();
});
更多信息:LINK
编辑:其他信息LINK2
答案 1 :(得分:0)
您需要先包含jquery。 然后尝试
$('#div1').mouseover(function() {
$('#div1').style.display='none';
$('#div2').style.display='block';
});
$('#div2').mouseout(function() {
$('#div1').style.display='block';
});
$('#btn').onclick(function() {
$('#div3').style.display='block';
});
我希望,这有效;) 您也可以尝试使用
$('#div1').style.visibility='hidden';
而不是
$('#div1').style.display='none';
答案 2 :(得分:0)
Pure Javascript中的东西:
<script>
var div_one = document.getElementById("1")
var div_two = document.getElementById('2');
var div_three = document.getElementById('3');
var button = document.getElementById('btn');
var common_div = 2;
div_one.onmouseover=function(){
if(common_div==3)
div_three.style.display = 'block';
else
div_two.style.display = 'block';
div_one.style.display = 'none';
};
div_two.onmouseout=function(){
div_one.style.display = 'block';
div_two.style.display = 'none';
};
div_three.onmouseout=function(){
common_div = 3;
div_one.style.display = 'block';
div_two.style.display = 'none';
div_three.style.display = 'none';
};
button.onclick = function(){
div_three.style.display = 'block';
}
</script>