所以我得到了多个嵌入不同图像的div。每个都有其唯一的名称属性。我正在尝试通过更改图像源将悬停效果应用于每个div。我不想编写多个脚本,而是尝试编写一个会影响每个div的脚本块。
<div id="div1" >
<img id="img1" name="img1" src="img1_up.jpg" />
</div>
<div id="div2">
<img id="img2" name="img2" src="img2_up.jpg" />
</div>...and so on
现在这是我目前拥有的用于翻转效果的脚本
<script>
var var1 = document.getElementById("div1");
var1.addEventListener("mouseover", changeImage1);
var1.addEventListener("mouseout", restoreImage1);
function changeImage1() {
document.getElementById("img1").src = "img1_ro.jpg";
}
function restoreImage1() {
document.getElementById("img1").src = "img1_up.jpg";
}
var var2 = document.getElementById("div2");
var2.addEventListener("mouseover", changeImage2);
var2.addEventListener("mouseout", restoreImage2);
function changeImage2() {
document.getElementById("img2").src = "img2_ro.jpg";
}
function restoreImage2() {
document.getElementById("img2").src = "img2_up.jpg";
}...and so on
</script>
我想使用每个图像的名称属性来创建应用于所有图像的动态代码。这是我的想法,但不确定确切的编写方式。请帮忙
...
var dynamicVar = ????
dynamicVar.addEventListener("mouseover", changeImage();
dynamicVar.addEventListener("mouseout", restoreImage();
function changeImage() {
document.getElementById(dynamicVar).src = dynamicVar + "_ro.jpg";
}
function restoreImage() {
document.getElementById(dynamicVar).src = dynamicVar + "_up.jpg";
}
您可以使用loop
添加事件,无需为每个div 指定 id:
var inputs = document.getElementsByTagName("div");
for(var i = 0; i < inputs.length; i++) {
if(inputs[i].id.indexOf('div') >= 0) {
inputs[i].addEventListener("mouseover", changeImage);
inputs[i].addEventListener("mouseout", restoreImage);
}
}
function changeImage(){
var tmpStr = this.id;
var divIndex = tmpStr.substring(3, tmpStr.length);
document.getElementById("img" + divIndex).src = divIndex + "_ro.jpg";
}
function restoreImage(){
var tmpStr = this.id;
var divIndex = tmpStr.substring(3, tmpStr.length);
document.getElementById("img" + divIndex).src = divIndex + "_up.jpg";
}
见小提琴:链接
试试这个
var parent = document.getElementById("parent");
var childs = parent.getElementsByTagName('div');
for (var i = 0; i < childs.length; i++) {
(function () {
var e = childs[i];
e.addEventListener("mouseover", function () {
changeImage(e);
});
e.addEventListener("mouseout", function () {
restoreImage(e);
});
}());
}
function changeImage(element) {
var imgs = element.getElementsByTagName('img');
for (var i = 0; i < imgs.length; i++) {
alert(imgs[i].id);
}
}
function restoreImage(element) {
var imgs = element.getElementsByTagName('img');
for (var i = 0; i < imgs.length; i++) {
imgs[i].src = img_ro;
}
}
你可以检查这个小提琴