0

我知道这已经被问了几十次了,但根据回复,我似乎是正确的。我有日历显示,但没有内容。这是我的代码:

<link rel='stylesheet' type='text/css' href="fullcalendar/fullcalendar.css" />
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript" src="fullcalendar/gcal.js"></script>
<script src="lib/jquery.min.js"></script>
<script src="lib/jquery-ui.custom.min.js"></script>
<script src="fullcalendar/fullcalendar.min.js"></script>

<script>
$(document).ready( function() {
$('#calendar').fullCalendar({
    events: { url: "https://www.google.com/calendar/feeds/cathysegalgarcia%40gmail.com/public/basic" 
    }
});
});
</script>

感谢帮助!

4

1 回答 1

0

我在演示版中尝试了您的网址,它有效。我会说检查你的脚本 srcs,看起来你有两个 jquery.min.js srcs。

我从演示 gcal.html 中尝试了这个,它工作得很好。

<!DOCTYPE html>
<html>
<head>
<link href='../fullcalendar/fullcalendar.css' rel='stylesheet' />
<link href='../fullcalendar/fullcalendar.print.css' rel='stylesheet' media='print' />
<script src='../lib/jquery.min.js'></script>
<script src='../lib/jquery-ui.custom.min.js'></script>
<script src='../fullcalendar/fullcalendar.min.js'></script>
<script src='../fullcalendar/gcal.js'></script>
<script>
    $(document).ready(function() {
        $('#calendar').fullCalendar({
            events: { url: "https://www.google.com/calendar/feeds/cathysegalgarcia%40gmail.com/public/basic"
            }
        });
    });
</script>
<style>

    body {
        margin-top: 40px;
        text-align: center;
        font-size: 14px;
        font-family: "Lucida Grande",Helvetica,Arial,Verdana,sans-serif;
        }

    #loading {
        position: absolute;
        top: 5px;
        right: 5px;
        }

    #calendar {
        width: 900px;
        margin: 0 auto;
        }

</style>
</head>
<body>
<div id='calendar'></div>
</body>
</html>
于 2013-10-29T00:56:37.663 回答