我已经记录了一个iMacro,它在两个下拉列表中处理设置日期,明天和后天。但是使用它的人必须每天编辑宏来设置日期,如果它明天或后天发生变化,包括月份。
我想知道imacro是否可以自动完成?将日期1设置为明天的日期,将日期2设置为后天的日期。那么一个月呢?如果从明天或后天开始新的一个月呢?imacro自己能做出一些常识性的决定吗?:)
发布于 2018-02-10 02:42:48
这段代码每天输出明天,后天输出。需要使用Javascript和iMacros命令EVAL。
var today = new Date();
var tomorrow = new Date();
tomorrow.setDate(today.getDate() + 1);
var day = tomorrow.getDate().toString();
if (day.length < 2) {day = "0" + day;}
var month = (tomorrow.getMonth() + 1).toString();
if (month.length < 2) {month = "0" + month;}
var year = tomorrow.getFullYear().toString();
var tomorrowDateString = day + "/" + month + "/" + year;
var afterTomorrow = new Date();
afterTomorrow.setDate(today.getDate() + 2);
var day = afterTomorrow.getDate().toString();
if (day.length < 2) {day = "0" + day;}
var month = (afterTomorrow.getMonth() + 1).toString();
if (month.length < 2) {month = "0" + month;}
var year = afterTomorrow.getFullYear().toString();
var afterTomorrowDateString = day + "/" + month + "/" + year;
alert('tomorrow: '+tomorrowDateString+' after tomorrow: '+afterTomorrowDateString);
对于iMacros,此代码将如下所示:
SET tomorrow EVAL("var today = new Date(); var tomorrow = new Date(); tomorrow.setDate(today.getDate() + 1); var day = tomorrow.getDate().toString(); if (day.length < 2) {day = \"0\" + day;} var month = (tomorrow.getMonth() + 1).toString(); if (month.length < 2) {month = \"0\" + month;} var year = tomorrow.getFullYear().toString(); var tomorrowDateString = day + \"/\" + month + \"/\" + year; tomorrowDateString;")
PROMPT {{tomorrow}}
SET afterTomorrow EVAL("var today = new Date(); var afterTomorrow = new Date(); afterTomorrow.setDate(today.getDate() + 2); var day = afterTomorrow.getDate().toString(); if (day.length < 2) {day = \"0\" + day;} var month = (afterTomorrow.getMonth() + 1).toString(); if (month.length < 2) {month = \"0\" + month;} var year = afterTomorrow.getFullYear().toString(); var afterTomorrowDateString = day + \"/\" + month + \"/\" + year; afterTomorrowDateString;")
PROMPT {{afterTomorrow}}https://stackoverflow.com/questions/28789269
复制相似问题