当有人试图在其日历上提交要保存到服务器的活动时,我收到错误消息。感谢您的帮助,谢谢您的时间!如果您需要更多具体信息,请告诉我。
UPDATE :当我通过fullcalendar中的drop功能将事件添加到日历中时,我自己从推送切换到数组时,它似乎正常但我遇到了该代码的问题所以我使用了来自fullcalendar的clientevents而现在我收到了这个错误。关于修复可能是什么的任何想法?
我收到以下错误:
未捕获的TypeError:无法读取属性' _calendar' D(moment.min.js:6)的未定义在e(jquery-1.11.3.min.js:5)处于Vb (vquery-1.11.3.min.js:5)在Vb(jquery-1.11.3.min.js:5)在Vb (jquery-1.11.3.min.js:5)在Function.m.param (jquery-1.11.3.min.js:5)在Function.ajax(jquery-1.11.3.min.js:5)
在对象。 (日历:514)在Function.each (jquery-1.11.3.min.js:2)在Object.success(日历:500)
companyCalendar.blade.php
var emailContainer = {};
emailContainer.email = email;
console.log("AJAX call here to submit dropped events as guest.");
$.ajax({
type: "POST",
url: '/partialAccountCheck',
data: emailContainer,
success: function (data) {
console.log('success, proceed with adding events to the company calendar');
$.each(newEvents, function (i, event) {
if (event.title !== 'undefined' && event.title !== null && event.title !== undefined) {
console.log(JSON.stringify(event));
event.start = moment(event.start).toDate();
event.end = moment(event.end).toDate();
event.start = formatDate(event.start) + ' ' + event.start.getHours() + ':' + event.start.getMinutes() + ':00';
event.end = formatDate(event.end) + ' ' + event.end.getHours() + ':' + event.end.getMinutes() + ':00';
console.log('event start is: ' + event.start);
console.log('event end is: ' + event.end);
event.identifier = <?php echo json_encode($companyIdentifier) ?>;
event.email = email;
event.services = event.title;
event.startAppointmentTime = event.start;
event.endAppointmentTime = event.end;
console.log("AJAX call here adding dropped appointments as guest.");
$.ajax({
type: "POST",
url: 'submitCalendarEvent',
data: event,
success: function (data) {
console.log('success');
},
complete: function (data) {
console.log(data);
}
});
} else {
console.log('exclude from submission');
}
});
},
complete: function (data) {
console.log(data);
}
});
答案 0 :(得分:4)
我解决了为event.start和event.end创建变量的问题。
start=moment(event.start).format('Y-MM-DD HH:mm:ss');
end=moment(event.end).format('Y-MM-DD HH:mm:ss');
$.ajax({
url:"insert.php",
type:"POST",
data:{title:event.title, start:start, end:end},
答案 1 :(得分:2)
您需要将日期从RotateRight()
格式转换为RotateLeft()
。
date