1

我正在尝试修改一个简单的小 AE 脚本，该脚本在源文本中显示星期几。我需要它在明天和后天显示——我可以通过使用 +1 和 +2 来做到这一点——但没有条件它会中断。我猜它是这样的；如果 d>6 则 d=0？不能完全正确地编写代码，非常感谢一些帮助！

d = new Date(Date(0));
var weekday=new Array(7);
weekday[0]="Sunday";
weekday[1]="Monday";
weekday[2]="Tuesday";
weekday[3]="Wednesday";
weekday[4]="Thursday";
weekday[5]="Friday";
weekday[6]="Saturday";

weekday[d.getDay()]
4

1 回答 1

0

[编辑：非常抱歉，@ndix，但以前的代码（已删除）完全有缺陷。底片将不起作用。使用下面的代码]

timeDirectionIndex = 0;//initial for today or future
daysToAdd = -15;//you can make this zero or any positive or negative integer
d = new Date(Date(0));
var weekday=new Array(7);
if (daysToAdd < 0) {timeDirectionIndex = 1;}

//if time is reversed, uses second column

weekday[0]=["Sunday","Saturday"][timeDirectionIndex];
weekday[1]=["Monday","Friday"][timeDirectionIndex];
weekday[2]=["Tuesday","Thursday"][timeDirectionIndex];
weekday[3]=["Wednesday","Wednesday"][timeDirectionIndex];
weekday[4]=["Thursday","Tuesday"][timeDirectionIndex];
weekday[5]=["Friday","Monday"][timeDirectionIndex];
weekday[6]=["Saturday","Sunday"][timeDirectionIndex];

//Math.abs makes it always positive; we rely on order of array now
if (daysToAdd < 0) {
   wd=weekday[ ( Math.abs(daysToAdd) +  3 + ( 3 - d.getDay())  )%7 ];
} else {
   wd=weekday[ (Math.abs(daysToAdd) +  (  d.getDay()   ) )%7 ];
}

alert(wd);

[以前的代码不起作用]

于 2013-10-23T01:41:44.713 回答