我在将事件保存到数据库时遇到问题。老实说,我不知道该怎么做。我能够在phpmyadmin中保存完整日历中的事件,但我的问题是如何从完整日历中保存事件。
<!DOCTYPE html>
<html>
<head>
<meta charset='utf-8' />
<link href='css/fullcalendar.css' rel='stylesheet' />
<link href='css/fullcalendar.print.css' rel='stylesheet' media='print' />
<script src='js/moment.min.js'></script>
<script src='js/jquery.min.js'></script>
<script src='js/fullcalendar.min.js'></script>
<script>
$(document).ready(function() {
var calendar = $('#calendar').fullCalendar({
editable: true,
events: "http://localhost/fullcalendar/events.php",
//what goes here? :(
});
});
</script>
<style>
body {
margin: 0;
padding: 0;
font-family: "Lucida Grande",Helvetica,Arial,Verdana,sans-serif;
font-size: 14px;
}
#calendar {
width: 900px;
margin: 40px auto;
}
</style>
</head>
<body>
<div id='calendar'></div>
</body>
</html>
答案 0 :(得分:1)
你必须进行ajax调用。
例如:
var title = event.title;
var start = event.start.format();
var end = (event.end == null) ? start : event.end.format();
$.ajax({
url: 'process.php',
data: 'type=newtitle&title='+title+'&start='+start+'&end='+end+'&eventid='+event.id,
type: 'POST',
dataType: 'json',
success: function(response){
if(response.status != 'success')
revertFunc();
},......
在你的php文件中,你可以使用每个对象:
$type = $_POST['type'];
if($type == 'newtitle')
{
$title = $_POST['title'];
$insert = mysqli_query($con,"INSERT INTO mytable(`title`) VALUES('$title')");
$lastid = mysqli_insert_id($con);
echo json_encode(array('status'=>'success','eventid'=>$lastid));
}
答案 1 :(得分:-1)
var s ="[{'title':1,'description':'Test1','start':'2014-09-15 16:00:00'}]";
var myObject = eval('(' + s + ')');
$('#CalContainer').fullCalendar(
{
defaultView : 'aWeek',
slotEventOverlap : false,
weekends : true,
allDaySlot : false,
firstDay:1,
minTime:"06:00:00",
maxTime:"24:00:00",
forceEventDuration:true,
defaultTimedEventDuration:"01:00:00",
contentHeight : "400px",
events:myObject,
eventClick: function(calEvent, jsEvent, view)
{
alert(calEvent);
}
});