java2个时间点的比较用于维护判断accesstoken时间是否相差2小时
2019/7/29 1:20:36 | 阅2558 | 来源:好空间网络 [打印] [关闭] |
package com.message; import java.text.ParseException; import java.text.SimpleDateFormat; import java.util.Date; public class getShiJianCha { public static double shijiancha(String strtime,String endtime){ String t1 = strtime; String t2 = endtime; Date d1 = null; Date d2 = null; SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); try { d1 = sdf.parse(strtime); d2 = sdf.parse(endtime); } catch (ParseException pe) { System.out.println(pe.getMessage()); } long dd1 = d1.getTime(); long dd2 = d2.getTime(); double hours = (double) (dd2 - dd1) / 3600 / 1000; return hours; } }
测试java类
SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");//设置日期格式 String xianshijian=df.format(new Date()); System.out.println("现在时间:"+xianshijian); String yuanshijian="2019-07-28 23:59:59"; System.out.println("原时间:"+yuanshijian); getShiJianCha shijian = new getShiJianCha(); double cha= shijian.shijiancha(yuanshijian, xianshijian); System.out.println("相差:"+cha); if(cha<1.9) { System.out.println("有效期内"); }else { System.out.println("过期了"); }