我目前正在为我公司的新主页制作一个图片滑块,而且无法弄清楚javascript部分中的错误。 滑块包含4张图像,一旦到达最后一张图像,就应该再次从第一张图像开始。问题是图像按“1-2-1-2-3-4-1-2-3-4 ......”的顺序显示
这里是代码:
$('.slide').first().addClass('active');
$('.slide').hide();
$('.active').show();
$('#next').on('click', nextSlide);
$('#prev').on('click', prevSlide);
// Auto slider
if (options.autoswitch === true) {
setInterval(nextSlide, options.autoswitch_speed);
}
function nextSlide() {
$('.active').removeClass('active').addClass('prevActive');
if ($('.prevActive').is(':last-child')) {
$('.slide').first().addClass('active');
} else {
$('.prevActive').next().addClass('active');
}
$('.prevActive').removeClass('prevActive');
$('.slide').fadeOut(options.speed);
$('.active').fadeIn(options.speed);
}
function prevSlide() {
$('.active').removeClass('active').addClass('prevActive');
if ($('.prevActive').is(':first-child')) {
$('.slide').last().addClass('active');
} else {
$('.prevActive').prev().addClass('active');
}
$('.prevActive').removeClass('prevActive');
$('.slide').fadeOut(options.speed);
$('.active').fadeIn(options.speed);
}
});
和相应的CSS:
#slider-container {
height: 600px;
margin: 0 auto;
max-width: 100%;
overflow: hidden;
position: relative;
width: 800px;
}
#sldier {
height: 100%;
width: 100%;
}
#slider .slide img {
height: 100%;
width: auto;
}
#prev, #next {
cursor: pointer;
max-width: 30px;
opacity: 1;
position: absolute;
top: 8%;
-webkit-transition: opacity 0.2s linear;
-moz-transition: opacity 0.2s linear;
-o-transition: opacity 0.2s linear;
transition: opacity 0.2s linear;
z-index: 999;
}
#prev { left: 12px; }
#next { right: 3px; }
#slider-container:hover #prev, #slider-container:hover #next { opacity: .7; }
.slide {
position: absolute;
width: 100%;
}
.slide-copy {
background: #777;
background: rgba(0, 0, 0, .5);
bottom: 0;
color: #fff;
left: 0;
padding: 20px;
position: absolute;
}
@media (min-width: 600px) {
#prev, #next {
top: 45%;
}
}
和HTML
<div id="slider-container">
<img src="img/arrowprev" id="prev" alt="prev">
<ul id="slider">
<li class="slide">
<div class="slide-copy">
<h2>Placeholder</h2>
<p>Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat.</p>
</div>
<img src="img/placeholder" alt="placeholder">
</li>
<li class="slide">
<div class="slide-copy">
<h2>Placeholder2</h2>
<p>Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat.</p>
</div>
<img src="img/placeholder" alt="placeholder">
</li>
<li class="slide">
<div class="slide-copy">
<h2>Placeholder3</h2>
<p>Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat.</p>
</div>
<img src="img/placeholder" alt="placeholder">
</li>
<li class="slide">
<div class="slide-copy">
<h2>Placeholder4</h2>
<p>Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat.</p>
</div>
<img src="img/placeholder" alt="placeholder">
</li>
</ul>
<img src="img/arrownext" id="next" alt="next">
</div>
可能与
有关if ($('.prevActive').is(':first-child')) {
$('.slide').last().addClass('active');
我无法弄清楚问题并感谢任何帮助。 提前谢谢你:)
这是我作为来源的页面: https://www.jqueryscript.net/slider/Tiny-jQuery-Image-Slider-Slideshow-With-Caption-Support.html
答案 0 :(得分:0)
我发现失败,它不起作用的独特之处在于对每个next / prev操作的每个prevActive
和active
元素应用可见性,因此我&#39; ve将这些行添加到两个函数中:
$('.prevActive').hide();
$('.active').show();
(我还评论了引用选项变量的行,如说明中未定义的那样)。
你可以在这里看到小提琴: https://jsfiddle.net/a4bssrf5/8/