嗨,我正在尝试使用CSS在屏幕上移动图像.目前,当我运行它时,它只显示在页面顶部.这是我正在尝试的代码.我试图在chrome中运行它.
<head>
<style>
#float{
width: 200px;
height: 500px;
position: relative;
animation: floatBubble 2s inifnate;
animation-directon: normal;
}
@keyframes floatBubble{
0% {
top:0;
-webkit-animation-timing-function: ease-in;
}
100% {
top: 500px;
animation-timing-function: ease-out;
}
}
</style>
</head>
<body style="background-color:#FF9900; color:#CC0033; text-align:center">
<div id="float">
<img src ="bub.png" height="100px" width="100px" alt="bubbles">
</div>
</body>
我正在创建一个脚本,从表单中获取信息并将其写入日历.当我在脚本编辑器中运行脚本时,它工作得很好,从电子表格中获取信息并创建日历事件.但是,当我填写格式并提交(提交是触发器)时,我收到一条错误,指出我没有权限调用getActiveForm().我拥有表格和日历.我很欣赏任何想法.
码
function createEvent() {
var calendarToUse = "testing";
var ssUrlToUse ='https://docs.google.com/a/bay.k12.fl.us/spreadsheets
/d/1ZTDQL9G5U7RqbbKQbAfb3ERqFwpSsC3EOQxdD1zdQwA/edit#gid=441997215';
/* Get Canalnder*/
var calen = CalendarApp.getCalendarsByName(calendarToUse);
var cal = calen[0];
Logger.log(cal.getName());
/* get info from responses*/
var mySS
=SpreadsheetApp.openByUrl(ssUrlToUse).getActiveSheet()
Logger.log(mySS.getName());
var values = mySS.getDataRange().getValues();
var response = values[values.length-1];
var i=2;
var desc = response[i];
var loc = response[i+1];
var start = makeGreatDate( response[i+2], response[i+4]);
var end = makeGreatDate(response[i+3],response[i+6]);
var title = response[i+5];
/* populate calendar event*/
var event = cal.createEvent(title, start, end, {
description : desc, …