javascript-未捕获的typeerror $(…).swipe不是函数
作者:互联网
我试图在博客中创建可滑动菜单,这是我的代码:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script type="text/javascript" src="https://www.googledrive.com/host/0B2Iusn9ixPQ2cFFldHoweXRsWms"></script>
<script type="text/javascript">
$(window).load(function(){
$("[data-toggle]").click(function() {
var toggle_el = $(this).data("toggle");
$(toggle_el).toggleClass("open-sidebar");
});
$(".swipe-area").swipe({
swipeStatus:function(event, phase, direction, distance, duration, fingers)
{
if (phase=="move" && direction =="right") {
$(".container").addClass("open-sidebar");
return false;
}
if (phase=="move" && direction =="left") {
$(".container").removeClass("open-sidebar");
return false;
}
}
});
});
</script>
并且Iam在“ $(”.swipe-area“).swipe({”:未捕获的typeerror $(…).swipe不是函数
请帮忙,
谢谢
解决方法:
尝试这个
<html>
<head>
<script src="http://code.jquery.com/jquery-1.11.3.min.js"></script>
<script src="http://code.jquery.com/mobile/1.4.5/jquery.mobile-1.4.5.min.js"></script>
<script>
$(document).on("pagecreate","#pageone",function(){
$("p").on("swipe",function(){
$(this).hide();
});
});
</script>
</head>
<body>
<div data-role="page" id="pageone">
<div data-role="main" class="ui-content">
<p>If you swipe me, I will disappear.</p>
<p>Swipe me away!</p>
<p>Swipe me too!</p>
</div>
</div>
</body>
</html>
更新-意外令牌
您的代码在下面的行中有语法错误
swipeStatus:function(event, phase, direction, distance, duration, fingers)
正确的语法-您可以通过以下方式定义swipeStatus函数
var swipeStatus = function(event, phase, direction, distance, duration, fingers)
标签:blogger,javascript,jquery 来源: https://codeday.me/bug/20191119/2039981.html