在小提琴上,我找到了一个简单的旋转器,试图让它在我死亡简单的HTML页面中运行。
页面示例如下:
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Demo</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<style>
img { max-height: 100px }
.rotator img { max-height: 200px; border: dashed 5px pink; }
</style>
<script>
$(document).ready(function() {
alert('aaa');
var $rotator = $(".rotator");
$rotator.find("img:gt(0)").hide();
setTimeout(Rotate, 1000);
function Rotate() {
var $current = $rotator.find("img:visible");
var $next = $current.next();
if ($next.length == 0) $next = $rotator.find("img:eq(0)");
$current.hide();
$next.show();
setTimeout(Rotate, 5000);
}
});
</script>
</head>
<body>
<img src="http://2.bp.blogspot.com/-XI9yzJrwLac/TkLKLZF_kDI/AAAAAAAACFE/PxPDRzwa4tQ/s1600/cute+cats+pictures+3.jpg"/>
<img src="http://2.bp.blogspot.com/-NOD8B0m7MEE/TrvJAVAPYWI/AAAAAAAAAuE/KoffoIdQfNk/s640/cute-kittens-in-cups-pics.jpg"/>
<img src="http://1.bp.blogspot.com/_cWcuJM9QIG4/S7rOVzM1YcI/AAAAAAAAAgQ/RJx5oR55Ekk/s640/Animal+wallpapers%252Bcat+wallpapers%252Bmobile+wallpapers%252Bpc+wallpapers%252Bmobile+themes%252Bpc+themes+15cc.jpg"/>
<div class="rotator">
<a href="http://google.com">
<img src="http://2.bp.blogspot.com/-XI9yzJrwLac/TkLKLZF_kDI/AAAAAAAACFE/PxPDRzwa4tQ/s1600/cute+cats+pictures+3.jpg"/>
</a>
<a href="http://google.com">
<img src="http://2.bp.blogspot.com/-NOD8B0m7MEE/TrvJAVAPYWI/AAAAAAAAAuE/KoffoIdQfNk/s640/cute-kittens-in-cups-pics.jpg"/>
<a>
<a href="http://google.com">
<img src="http://1.bp.blogspot.com/_cWcuJM9QIG4/S7rOVzM1YcI/AAAAAAAAAgQ/RJx5oR55Ekk/s640/Animal+wallpapers%252Bcat+wallpapers%252Bmobile+wallpapers%252Bpc+wallpapers%252Bmobile+themes%252Bpc+themes+15cc.jpg"/></a>
</div>
<label />
</body>
</html>
简单的脚本应该定期切换图像,但不是只显示所有3个图像。并且不显示警报消息。
我试图调试代码,当我删除函数Rotate()
时,页面上会出现一条警告消息。
为什么函数Rotate()
不起作用?
答案 0 :(得分:1)
$.next()
获取集合中的直接元素。您的设置仅包含可见图像 - 即仅包含一个。怎么会有下一个元素?
工作小提琴:http://jsfiddle.net/gjzd7/
我所做的只是将$.next()
调用更改为$.index()
请求,并按照图像数量对其进行模块化(因此您永远不会获得不存在的图像)。如果您需要修改其他任何内容或任何解释,请告诉我们!
答案 1 :(得分:1)
您还可以将图像缓存在jQuery数组中,并按如下方式迭代它们。
var imgs = $(".slides"); // images to be rotated
var current = 0;
function rotate( ) {
// set current to next image
current = current >= imgs.length ? 0 : current + 1;
$(".rotator").prop("src", $(imgs.get(current)).prop("src") );
setTimeout( rotate, 5000 );
}
rotate();
示例here
答案 2 :(得分:0)
使用此简单脚本使用按钮
旋转图像function rotateImage(degree) {
$('#demo-image').animate({ transform: degree }, {
step: function(now,fx) {
$(this).css({
'-webkit-transform':'rotate('+now+'deg)',
'-moz-transform':'rotate('+now+'deg)',
'transform':'rotate('+now+'deg)'
});
}
});
}
<style>
#demo-image{padding:35px 15px;}
.btnRotate {padding: 15px 20px;border:1px solid #000; background-color:#999;color: #000;cursor: pointer;}
</style>
<div style="height:600px; width:800px; background-color:#CCC; border:1px solid #000; border-radius:6px; margin-left:auto; margin-right:auto;">
<div style="width:550px; margin-left:auto; margin-right:auto;">
<label>Rotate Image:</label>
<input type="button" class="btnRotate" value="30" onClick="rotateImage(this.value);" />
<input type="button" class="btnRotate" value="60" onClick="rotateImage(this.value);" />
<input type="button" class="btnRotate" value="90"onClick="rotateImage(this.value);" />
<input type="button" class="btnRotate" value="180" onClick="rotateImage(this.value);" />
<input type="button" class="btnRotate" value="-180" onClick="rotateImage(this.value);" />
<input type="button" class="btnRotate" value="360" onClick="rotateImage(this.value);" />
</div>
<div style="width:350px; margin-left:auto; margin-right:auto; margin-top:25px;"><img src="image-rotating-script-using-jquery .jpg" id="demo-image" /></div>
</div>