Question

我在jquery中构建了一个简单的滑块。它工作正常。但现在我想根据图像的变化突出显示滑块底部的方块。

我的代码就在这里。

我使用了以下的html代码。

<div>
<div class="slider">
<span class="control_next">Next</span>
<span class="control_prev">Previous</span>
<ul>
<li style="background:#ccc;">Image 1</li>
<li style="background:#547;">Image 2</li>
<li style="background:#124;">Image 3</li>
<li style="background:#054;">Image 4</li>
</ul>
</div>
<div class="square square1"></div>
<div class="square square2"></div>
<div class="square square3"></div>
<div class="square square4"></div>
</div>

上述代码的css

.slider {
  position: relative;
  overflow: hidden;
  border: 1px solid #ccc;
}

.slider ul {
  position: relative;
  margin: 0;
  padding: 0;
  list-style: none;
}

.slider ul li {
  position: relative;
  display: block;
  float: left;
  margin: 0px;
  padding: 0;
  width: 500px;
  height: 150px;
  color

}

.control_prev, .control_next {
 position: absolute;
  bottom: 0;
  z-index: 999;
  display: block;
  padding: 5px;
  width: auto;
  height: auto;
  background: #2a2a2a;
  color: #fff;
  text-decoration: none;
  font-size: 18px;
  opacity: 0.8;
  cursor: pointer;
  margin:5px;
}

.control_prev:hover, .control_next:hover {
  opacity: 1;
  -webkit-transition: all 0.2s ease;
}

.control_prev {
  right:50;
  border-radius: 0 2px 2px 0;
}
.control_next {
  right: 0;
  border-radius: 2px 0 0 2px;
}
.square
{
  z-index: 999;
  height:20px;
  width:20px;
  background:#ccc;
  position: relative;
  float: left;
  margin:0px 5px; 
}

和用于滑动图像的jquery是

$(document).ready(function(){
var slideCount = $('.slider ul li').length;
var slideWidth = $('.slider ul li').width();
var slideHeight = $('.slider ul li').height();
var sliderUlWidth = slideCount * slideWidth;
$('.slider').css({ width: slideWidth, height: + slideHeight });

$('.slider ul').css({ width: sliderUlWidth, marginLeft: - slideWidth });

$('.slider ul li:last-child').prependTo('.slider ul');
function moveLeft() {
    $('.slider ul').animate({
        left: + slideWidth
    }, 700, function () {
        $('.slider ul li:last-child').prependTo('.slider ul');
        $('.slider ul').css('left', '');
    });
};

 function moveRight() {
    $('.slider ul').animate({
        left: - slideWidth
    }, 700, function () {
        $('.slider ul li:first-child').appendTo('.slider ul');
        $('.slider ul').css('left', '');
    });
};

$('.control_prev').click(function () {
    moveLeft();
});

$('.control_next').click(function () {
    moveRight();
});

$(function(){
setInterval(function () {
    moveRight();
}, 5000);
});

});

请帮我完成此操作。感谢。

Answer 1

您需要跟踪当前显示的图像，然后相应地向“正方形”添加一个类：http://jsfiddle.net/WZCA3/

//additional variables
var allSquares = $('.square');
var totalSquares = allSquares.length;
var index = 0;

//the move function
function moveRight() {
    index++;
    $('.slider ul').animate({
        left: - slideWidth
    }, 700, function () {
        $('.slider ul li:first-child').appendTo('.slider ul');
        $('.slider ul').css('left', '');
    });
     setSquare();
};

// the active status
function setSquare() {
    allSquares.removeClass("active").eq(index % totalSquares).addClass("active");   
}

按照jsfiddle链接查看整个代码。

索引变量基本上跟踪当前显示的图像。当滑块转到下一个图像时它会增加1，当它转到前一个图像时它会减1。每次更改图像时，都会获得当前索引，并通过生成currentIndex％numberOfImages找到它对应的“square”，在您的情况下，这将始终是0到3之间的值，始终设置正确的方块

如何使用方形项目符号构建JQuery滑块？

1 个答案: