我使用jQuery插件fullcalendar写了一个代码,一切正常,直到我在eventClick中发送$ .ajax请求。
我试图提醒事件中的某些内容并且它有效,但是ajax请求不起作用。 js代码:
!function($)
{
$(document).ready(function() {
$("#eventDialog").dialog({
autoOpen : false,
modal : true
});
var date = new Date();
var d = date.getDay();
var m = date.getMonth();
var y = date.getFullYear();
$("#calendar").fullCalendar({
theme : true,
header : {
left : 'next,prev today',
center : 'title',
right : 'month,agendaWeek,agendaDay'
},
editable : false,
events : [
<?php
while ($event = mysql_fetch_array($selectevents))
{
$startd = explode(".", $event['start']);
$endd = explode(".", $event['end']);
$starth = explode(":", $event['starth']);
$endh = explode(":", $event['endh']);
?>
{
id : <?php echo $event['id']; ?>,
title : '<?php echo stripslashes($event['title']); ?>',
<?php
if ($event['allday'] == 1)
{
?>
start : new Date(<?php echo $startd[2]; ?>,<?php echo $startd[1]-1; ?>,<?php echo $startd[0]; ?>),
end : new Date(<?php echo $endd[2]; ?>,<?php echo $endd[1]-1; ?>,<?php echo $endd[0]; ?>)
<?php
}
else
{
?>
start : new Date(<?php echo $startd[2]; ?>,<?php echo $startd[1]-1; ?>,<?php echo $startd[0]; ?>,<?php echo $starth[0]; ?>,<?php echo $starth[1]; ?>),
end : new Date(<?php echo $endd[2]; ?>,<?php echo $endd[1]-1; ?>,<?php echo $endd[0]; ?>,<?php echo $endh[0]; ?>,<?php echo $endh[1]; ?>),
allDay : false
<?php
}
?>
}
<?php
$evNum--;
if ($evNum > 0)
echo ",";
}
?>
],
eventClick : function(event) {
var eid = event.id;
$.ajax({
type : "GET",
url : "getEvent.php",
data : "id=" + eid,
succuss : function(msg) {
var title = $("event title", msg).text();
var description = $("event description", msg).text();
var start = $("event start", msg).text();
var end = $("event end", msg).text();
var starth = $("event starth", msg).text();
var endh = $("event endh", msg).text();
$("#eventDialog").attr("title", title);
$("#eventDialog #edDescription").html(title);
eventDialog.dialog("open");
}
});
}
});
});
}(jQuery);
getEvent.php
$id = htmlspecialchars(mysql_real_escape_string($_GET['id']));
$selectevent = mysql_query("SELECT * FROM `calendar` WHERE `id`='$id'") or die(mysql_error());
$event = mysql_fetch_array($selectevent);
header("Content-Type: text/xml; charset=utf-8");
echo '<?xml version="1.0" encoding="utf-8" ?>';
echo "<event>";
echo "<id>".$event['id']."</id>";
echo "<title>".stripslashes($event['title'])."</title>";
echo "<description>".$event['description']."</description>";
echo "<start>".$event['start']."</start>";
echo "<end>".$event['end']."</end>";
echo "<starth>".$event['starth']."</starth>";
echo "<endh>".$event['endh']."</endh>";
echo "</event>";
有人可以指出问题吗?
感谢的!
答案 0 :(得分:1)
多汁?,我猜这是你的问题,应该是“成功”
答案 1 :(得分:1)
在您的代码中将succuss
更改为success
,它必须是拼写错误。
由于jQuery
$.ajax
在其设置中未找到success
处理程序,因此无法执行任何操作。