【发布时间】:2021-09-14 23:19:46
【问题描述】:
我偶然发现了一个问题,我有一个 Jquery 脚本来更改样式表,原始脚本使用一个按钮。我想使用带有 Foundation css 框架的开关,但我无法让它工作,也许我错过了一些东西。这里是 jquery 代码:
var click = false;
$("#Switch").on("click", function () {
if (!click) {
$('link[href*="<?php echo
base_url('/Foundation/assets/css/dark/foundation.css')?>"]').attr(
"href",
"<?php echo
base_url('/Foundation/assets/css/dark/foundation.css')?>"
);
click = true;
console.log("changed to style1.css");
} else {
$('link[href*="<?php echo
base_url('/Foundation/assets/css/lumen/foundation.css')?>"]').attr(
"href",
"<?php echo
base_url('/Foundation/assets/css/lumen/foundation.css')?>"
);
click = false;
console.log("changed to style.css");
}
});
这里是 header.php 代码:
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<link rel="stylesheet" href="<?php echo
'base_url'('Foundation/assets/css/foundation.css'); ?>">
<link rel="stylesheet" href="<?php echo
'base_url'('Foundation/assets/css/main.css'); ?>">
<link rel="stylesheet" href="<?php echo
'base_url'('FontAwesome/css/fontawesome.css'); ?>">
<title><?php echo $title ?></title>
</head>
这里是开关:
<input
class="switch-input Switch"
id="Switch"
type="checkbox"
name="exampleSwitch"
/>
<label class="switch-paddle" for="Switch">
<span class="show-for-sr">Download Kittens</span>
</label>
【问题讨论】:
标签: jquery css codeigniter zurb-foundation