$(document).ready(function() {
//set all logo images based on data returned from database
var WinBack = $('div.jWinBackFrom').html();
switch(WinBack) {
case '1': //If database returns 1 : set Verizon FiOS logo
$('div.jWinBackFrom').prepend('<img id="WinBackLogo" src="<?php echo site_url();?>/assets/images/verizon-fios-logo.jpg" />');
break;
case '2': //If database returns 2 : set DirecTV logo
$('div.jWinBackFrom').prepend('<img id="WinBackLogo" src="<?php echo base_url();?>assets/images/directv-logo.jpg" />');
break;
case '3': //If database returns 3 : set DISH Network logo
$('div.jWinBackFrom').prepend('<img id="WinBackLogo" src="<?php echo base_url();?>assets/images/dish-network-logo.jpg" />');
break;
}
在页面中,我有一个for循环内部有多个
<div class="jWinBackFrom"><?php echo $optimum['cWinBackFrom'][$i]; ?></div>
。想象一下输出是这样的:
<div class="jWinBackFrom">1</div>
,<div class="jWinBackFrom">2</div>
,<div class="jWinBackFrom">3</div>
,<div class="jWinBackFrom">2</div>
,<div class="jWinBackFrom">1</div>``<div class="jWinBackFrom">3</div>
,<div class="jWinBackFrom">1</div>
。基本上都是随机的。jQuery需要从所有这些DIV获取值,并将数字值与相关联的图像交换。请问我这样做是否正确(意思是使用了正确的方法),这是最有效的方法吗?如果不是,那什么方法更好呢?
我需要为许多其他字段重复此过程。我的理论是,让数据库返回简单的值,如1,并使用jQuery动态构建页面速度更快。
感谢帮助!谢谢!