Javascript 鼠标悬停-使图像显示
当我将鼠标移到某个图像贴图上时,我试图使图像出现,但我的其他mouseover和mouseout功能仍在工作Javascript 鼠标悬停-使图像显示,javascript,jquery,html,css,image,Javascript,Jquery,Html,Css,Image,当我将鼠标移到某个图像贴图上时,我试图使图像出现,但我的其他mouseover和mouseout功能仍在工作 <script> var img1 = document.getElementById("img1"), img2 = document.getElementById("img2"); img2.style.display = "none"; // hide when page is loaded img1.onmouseover = funct
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>
我正在使用:
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>
基本上是这样
<style>
#img2{
position: fixed;
right:0;
bottom:0;
}
</style>
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>
#img2{
位置:固定;
右:0;
底部:0;
}
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>
var img1=document.getElementById(“img1”),
img2=document.getElementById(“img2”);
img2.style.display=“无”//加载页面时隐藏
img1.onmouseover=函数(){
img2.style.display=“块”;
}
img1.onmouseout=函数(){
img2.style.display=“无”;
}
我现在在一个网站上使用这种方法。但是我在地图区域内使用的不是
,而是一个ID。像这样
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>
<map name="map12" id="img_id12">
<area id="img2" class="youtube" coords="3878,24,3957,96" shape="rect" href="https://www.youtube.com/embed/skV-q5KjrUA" style="outline:none;"
onmouseover="if(document.images) document.getElementById('img_id12').src= 'assets/images/text/day/12solaire-ani.gif'; PlaySound('solaire'); "
onmouseout="if(document.images) document.getElementById('img_id12').src= 'assets/images/no-text/day/12.gif'; StopSound('solaire'); PlaySound('solaire-stop');" />
</map>
这个结果是当我将鼠标悬停在图像地图上时显示的图像。但是,它取消了我的其他功能。如何让所有东西协同工作?最好使用CSS:hover伪类来显示img2
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>
去掉img1.onmouseover和img1.onmouseout,并在样式块中添加以下css:
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>
#img1:hover + #img2 {display:block}
我在这里创建了一个小提琴示例 除了onmouseover和onmouseout之外,我没有看到任何其他函数。还取消了什么?请提供完整代码,然后我会检查它发生了什么@StephanieHallberg?图像是以编程方式生成的,这就是样式表方法无法工作的原因吗?你能用你的代码创建一个JSFIDLE吗?你看过我的小提琴了吗?
<script>
var img1 = document.getElementById("img1"),
img2 = document.getElementById("img2");
img2.style.display = "none"; // hide when page is loaded
img1.onmouseover = function(){
img2.style.display = "block";
}
img1.onmouseout = function(){
img2.style.display = "none";
}
</script>