一、简介

有个需求是，累加通话时长，把上次通话时长加上本次通话时长，依次累加。

二、实现

@Test

public void TestA() throws ParseException {

SimpleDateFormat myFormatter = new SimpleDateFormat("HH:mm:ss");

SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

//上一次通话时长

HealthConsultation healthConsultation = new HealthConsultation();

healthConsultation.setConsultinghours("00:00:08");

//呼叫时长 3

HealthCallStateRecord healthCallStateRecorda = new HealthCallStateRecord();

healthCallStateRecorda.setCreatedate(format.parse("2017-12-19 17:43:42"));

//本次通话时长 4

HealthCallStateRecord healthCallStateRecordb = new HealthCallStateRecord();

healthCallStateRecordb.setCreatedate(format.parse("2017-12-19 18:51:50"));

//4-3的时间 获得两个时间的毫秒时间差异

long diff = healthCallStateRecordb.getCreatedate().getTime() - healthCallStateRecorda.getCreatedate().getTime();//这样得到的差值是微秒级别

long nd = 1000 * 24 * 60 * 60;

long nh = 1000 * 60 * 60;

long nm = 1000 * 60;

long ns = 1000;

// 计算差多少天

long day = diff / nd;

// 计算差多少小时

long hour = diff % nd / nh;

// 计算差多少分钟

long min = diff % nd % nh / nm;

// 计算差多少秒//输出结果

long sec = diff % nd % nh % nm / ns;

//判断上次是否有通话，如果有通话时长，就都累加

if(healthConsultation.getConsultinghours() != null){

Calendar cal = Calendar.getInstance();

cal.setTime(myFormatter.parse(healthConsultation.getConsultinghours()));

int shi = (int)cal.get(Calendar.HOUR_OF_DAY)+(int)hour;

int fendo = (int)cal.get(Calendar.MINUTE)+(int)min;

int miao = (int)cal.get(Calendar.SECOND)+(int)sec;

//秒大于60设为00,然后判断分是否大于60分钟,如果是则分钟清0,小时加1,否则，分钟加1。

if(miao>60){

miao=00;

if(fendo>60){

fendo = 00;

shi = shi+1;

}else {

fendo = fendo + 1;

}

}

Date dates = myFormatter.parse(shi + ":" + fendo + ":" + miao);

System.out.println("不为空当前时间 : " + myFormatter.format(dates));

}else {

//如果上次无通话，就把本次时长，设为通话时长。

Date dates = myFormatter.parse(hour + ":" + min + ":" + sec);

System.out.println("为空当前时间 : " + myFormatter.format(dates));

}

}输出结果为:

不为空当前时间 : 01:08:16通话时长为1个多小时。