令人尴尬的是,我需要一个小算法
我正在帮助一个网站,其中gif B显示在7月1日,C在3和A在5。然后在第七天再次更改b,依此类推- a,b,c每隔一天上午8点更改一次
那么,我如何使用一个简单的算法来使此脚本返回0、1或2,具体取决于自7月1日以来的多少天
http://jsfiddle.net/mplungjan/T2Tjh/
var startDate = new Date(2012,6,1,8,0,0);
var pics = ["b.gif","c.gif","a.gif"];
var aDay = 24*60*60*1000;
function getDateDiff(startDate,endDate) {
var diff = endDate.getTime() - startDate.getTime();
return Math.round(diff/aDay);
}
document.write('<br/>'+getDateDiff(startDate,new Date()));发布于 2012-07-12 03:01:09
试试这个:
var startDate = new Date(2012, 6, 1, 8, 0, 0);
var pics = ["b.gif", "c.gif", "a.gif"];
var aDay = 24 * 60 * 60 * 1000;
function getDateDiff(startDate, endDate) {
var diff = endDate.getTime() - startDate.getTime();
return Math.round(diff / aDay);
}
document.write('<br/>' + pics[Math.floor(getDateDiff(startDate, new Date()) / 2) % 3]);发布于 2012-07-12 03:01:09
您将在此处使用modulo operator:
document.write('<br/>'+getDateDiff(startDate,new Date())%3);https://stackoverflow.com/questions/11439542
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