PHP获取指定时间段之间的 年,月,天,时,分,秒

核心代码：

Class Utils {
     
	public function fmDatetime($datetime) {
	  $year = substr($datetime,0,4);
	  $month = substr($datetime,5,2);
	  $day = substr($datetime,8,2);
	  $hour = substr($datetime,11,2);
	  $min = substr($datetime,14,2);
	  $sec = substr($datetime,17,2);
	  return mktime($hour,$min,$sec,$month,$day,0+$year);
	}
	
	 private function diffDateTime($DateStart,$DateEnd){
		$rs = array();
		
		$sYear = substr($DateStart,0,4);
		$eYear = substr($DateEnd,0,4);
		
		$sMonth = substr($DateStart,5,2);
		$eMonth = substr($DateEnd,5,2);
		
		$sDay = substr($DateStart,8,2);
		$eDay = substr($DateEnd,8,2);
		
		$startTime = $this->fmDatetime($DateStart);
		$endTime = $this->fmDatetime($DateEnd);
		$dis = $endTime-$startTime;//得到俩个时间的秒数
		$d = ceil($dis/(24*60*60));//得到天数
		$rs['day'] = $d;//天数
		$rs['hour'] = ceil($dis/(60*60));//小时
		$rs['minute'] = ceil($dis/60);//分钟
		$rs['second'] = $dis;//秒数
		$rs['week'] = ceil($d/7);//周
		
		$tem = ($eYear-$sYear)*12;//月份
		$tem1 = $eYear-$sYear;//年
		if($eMonth-$sMonth<0){//月份相减为负
			$tem +=($eMonth-$sMonth);
		}else if($eMonth==$sMonth){//月份相同
			if($eDay-$sDay>=0){
				$tem ++;
				$tem1++;
			}
		}else if($eMonth-$sMonth>0){//月份相减正负
			$tem1++;
			if($eDay-$sDay>=0){//且日期相减为正数
				$tem +=($eMonth-$sMonth)+1;
			}else{
				$tem +=($eMonth-$sMonth);
			}
		}
		$rs['month'] = $tem;
		$rs['year'] = $tem1;
		
		return $rs;
	}
}

一年多一天，返回的是2年，一个月多一天返回的是2个月，以此推......项目需要，才做此出来，开始我也到网上找这样的例子，但大家都是把年就按365天来算，月就按30天来算，这样算出来的结果肯定是没用的，年有可能是366天，月有可能是31，29，28都有可能