php – 从数据库中获取数据后如何在div中滑动
发布时间：2020-12-13 22:26:17 所属栏目：PHP教程 来源：网络整理
导读：我试图找到如何从数据库中提取数据后滑动数据块. 我尝试了很多,但没有得到结果.请帮助我找出解决方案 这是我的代码 htmlhead/headbody?php $servername="localhost";$username="root";$password="";$db="lalcoresidency";$conn=mysqli_connect($servername,$
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我试图找到如何从数据库中提取数据后滑动数据块.
我尝试了很多,但没有得到结果.请帮助我找出解决方案 这是我的代码 <html>
<head>
</head>
<body>
<?php
$servername="localhost";
$username="root";
$password="";
$db="lalcoresidency";
$conn=mysqli_connect($servername,$username,$password,$db);
if(!$conn){
die("connection failed:".mysqli_connect_error());
}
?>
<div class="flexslider">
<ul class="slides">
<?php
$query="select * from testimonial order by r_id desc";
$result = mysqli_query($conn,$query);
while($row=mysqli_fetch_array($result)){
?>
<li>
<?php echo $row['review'];?>
</li>
<?php
}
?>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js"></script>
<script>
$(document).ready(function() {
$('.flexslider').carousel({
interval: 3200
});
});
</script>
</body>
它已经花了我的时间,但我没有得到解决方案.这个代码只能逐行显示数据 my code output "Loving Staff Welcoming and helpful... Kind and welcome the food is great. Breakfast included." "Total value for money." "Excellent facilities with good location and very co-operative and efficient staff." "Accommodation worth the money paid for... Staff were very helpful and in-house car hire rates were so reasonable." "An excellent stay... Loved the stay and definitely look forward to keep going back for our next stay." 它一个接一个地显示数据.但我想显示第一行数据,然后滑动后需要显示第二行.然后滑动第三行. 解决方法
那是因为.carousel没有在jQuery中定义.
它是jQuery的插件,您可以从他们的网站下载. https://plugins.jquery.com/jcarousel/ 另外别忘了关闭< div>和< ul>标签. （编辑：李大同） 【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容! |