嗨，我是一个新的代码编码器，我想通过ID数据库进入jquery，但我不能这样做，我尝试搜索谷歌youtube，但我仍然不能这个代码我已经卡了几天了，谁能帮帮我呢?
这是原始代码:
<style>
@import url(http://fonts.googleapis.com /css?family=Open+Sans:400,300,600);
html {
border-top: 5px solid #fff;
background: #58DDAF;
color: #2a2a2a;
}
html, body {
margin: 0;
padding: 0;
font-family: 'Open Sans';
}
h1 {
color: #fff;
text-align: center;
font-weight: 300;
}
.slider {
position: relative;
overflow: hidden;
margin: 20px auto 0 auto;
border-radius: 4px;
}
.slider ul {
position: relative;
margin: 0;
padding: 0;
height: 200px;
list-style: none;
}
.slider ul li {
position: relative;
display: block;
float: left;
margin: 0;
padding: 0;
width: 500px;
height: 300px;
background: #ccc;
text-align: center;
line-height: 300px;
}
.slider ul li img{
width: 100%;
}
a#control_prev, a#control_next {
position: absolute;
top: 40%;
z-index: 999;
display: block;
padding: 4% 3%;
width: auto;
height: auto;
background: #2a2a2a;
color: #fff;
text-decoration: none;
font-weight: 600;
font-size: 18px;
opacity: 0.8;
cursor: pointer;
}
a#control_prev:hover, a#control_next:hover {
opacity: 1;
-webkit-transition: all 0.2s ease;
}
a#control_prev {
border-radius: 0 2px 2px 0;
}
a#control_next {
right: 0;
border-radius: 2px 0 0 2px;
}
.slider_option {
position: relative;
margin: 10px auto;
width: 160px;
font-size: 18px;
}
</style>
<script src="<?php echo base_url().'./style/jquery.js'; ?>"></script>
</head>
<body>
<script>
jQuery(document).ready(function ($) {
var slideCount = $('#slider2 ul li').length;
var slideWidth = $('#slider2 ul li').width();
var slideHeight = $('#slider2 ul li').height();
var sliderUlWidth = slideCount * slideWidth;
$('#slider2').css({ width: slideWidth, height: slideHeight });
$('#slider2 ul').css({ width: sliderUlWidth, marginLeft: - slideWidth });
$('#slider2 ul li:last-child').prependTo('#slider2 ul');
function moveLeft() {
$('#slider2 ul').animate({
left: + slideWidth
}, 200, function () {
$('#slider2 ul li:last-child').prependTo('#slider2 ul');
$('#slider2 ul').css('left', '');
});
};
function moveRight() {
$('#slider2 ul').animate({
left: - slideWidth
}, 200, function () {
$('#slider2 ul li:first-child').appendTo('#slider2 ul');
$('#slider2 ul').css('left', '');
});
};
$('a.control_prev1').click(function () {
moveLeft();
});
$('a.control_next1').click(function () {
moveRight();
});
});
</script>
<div id="slider2" class="slider">
<a href="#" id="control_next" class="control_next1">>></a>
<a href="#" id="control_prev" class="control_prev1"><</a>
<ul>
<li><img src="<?php echo base_url().'./images/events/ll.png' ?>"></li>
<li style="background: #aaa;"><img src="<?php echo base_url().'./images/events/blue.jpg' ?>"></li>
<li style="background: #aaa;"><img src="<?php echo base_url().'./images/events/red.jpg' ?>"></li>
</ul>
</div>
</body>
</html>
我试着传递ID，但它不工作，请帮助我:
My database: events
id image image2 image3
1 1a.jpg 2a.jpg 3a.jpg
2 1b.jpg 2b.jpg 3b.jpg
3 1b.jpg 2c.jpg 3c.jpg
<?php $query = $this->db->get('events'); ?>
<?php foreach($query->result() as $val): ?>
<script>
jQuery(document).ready(function ($) {
var id = "#slider<?php echo $val->id; ?>";
var ida = ".control_next<?php echo $val->id; ?>";
var idb = ".control_prev<?php echo $val->id; ?>";
var slideCount = $(id + ' ul li').length;
var slideWidth = $(id + ' ul li').width();
var slideHeight = $(id + ' ul li').height();
var sliderUlWidth = slideCount * slideWidth;
$(id).css({ width: slideWidth, height: slideHeight });
$(id + ' ul').css({ width: sliderUlWidth, marginLeft: - slideWidth });
$(id + ' ul li:last-child').prependTo(id + ' ul');
function moveLeft() {
$(id + ' ul').animate({
left: + slideWidth
}, 200, function () {
$(id + ' ul li:last-child').prependTo(id + ' ul');
$(id + ' ul').css('left', '');
});
};
function moveRight() {
$(id + ' ul').animate({
left: - slideWidth
}, 200, function () {
$(id + ' ul li:first-child').appendTo(id + ' ul');
$(id + ' ul').css('left', '');
});
};
$('a.control_prev ' + idb).click(function () {
moveLeft();
});
$('a.control_next ' + ida).click(function () {
moveRight();
});
});
</script>
<div id="slider<?php echo $val->id; ?>" class="slider">
<a href="#" id="control_next" class="control_next<?php echo $val->id; ?>">>></a>
<a href="#" id="control_prev" class="control_prev<?php echo $val->id; ?>"><</a>
<ul>
<li><img src="<?php echo base_url().'./images/events/'.$val->image; ?>"></li>
<li style="background: #aaa;"><img src="<?php echo base_url().'./images/events/'.$val->image2; ?>"></li>
<li style="background: #aaa;"><img src="<?php echo base_url().'./images/events/'.$val->image3; ?>"></li>
</ul>
</div>
<?php endforeach; ?>
无论何时使用id，都应该使用连接操作符。同时去掉id的括号。没必要。
var slideCount = $(id + ' ul li').length;
^ ^// add a space here.
var slideWidth = $(id + 'ul li').width();
var slideHeight = $(id + 'ul li').height();
我只给了3行，请改变它在所有需要的地方在你的jquery
$('a.' + idb).click(function () {
moveLeft();
});
$('a.' + ida).click(function () {
moveRight();
});