JAVA判断时间是否在时间区间内

package com.liying.tiger.test;

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;

public class Test {
    public static void main(String[] args) throws ParseException {
        String format = "HH:mm:ss";
        Date nowTime = new SimpleDateFormat(format).parse("09:27:00");
        Date startTime = new SimpleDateFormat(format).parse("09:27:00");
        Date endTime = new SimpleDateFormat(format).parse("09:27:59");
        System.out.println(isEffectiveDate(nowTime, startTime, endTime));
    }

    /**
     * 判断当前时间是否在[startTime, endTime]区间,注意时间格式要一致
     * 
     * @param nowTime 当前时间
     * @param startTime 开始时间
     * @param endTime 结束时间
     * @return
     * @author jqlin
     */
    public static boolean isEffectiveDate(Date nowTime, Date startTime, Date endTime) {
        if (nowTime.getTime() == startTime.getTime()
                || nowTime.getTime() == endTime.getTime()) {
            return true;
        }

        Calendar date = Calendar.getInstance();
        date.setTime(nowTime);

        Calendar begin = Calendar.getInstance();
        begin.setTime(startTime);

        Calendar end = Calendar.getInstance();
        end.setTime(endTime);

        if (date.after(begin) && date.before(end)) {
            return true;
        } else {
            return false;
        }
    }
}

 JAVA8:

//构造时间  00:20:55
        LocalTime time1 = LocalTime.of(8,20,55);
        LocalTime time3 = LocalTime.of(15,20,55);
        LocalTime time4 = LocalTime.of(18,10,55);

        if (time1.isBefore(time3) && time4.isAfter(time3)){
            System.out.println("在时间区间之内!");
        }

 

posted @ 2018-10-29 14:54  huanghaunghui  阅读(6390)  评论(0编辑  收藏  举报