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HomeBackend DevelopmentPHP Tutorial PHP获取指定时间段其间的 年,月,天,时,分,秒

PHP获取指定时间段之间的 年,月,天,时,分,秒
需求:前端传俩个标准的 时间格式过来,格式像2009-05-12 12:12:30,然后根据需要返回这个时间段的不同单位的表示
对于时间格式的校验我这里代码没贴出来,所以用的时候自己考虑加上

Class Utils {
         /**
	 * format MySQL DateTime (YYYY-MM-DD hh:mm:ss) 把mysql中查找出来的数据格式转换成时间秒数
	 * @param string $datetime
	 */
	public function fmDatetime($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);
	}
	/**
	 * 
	 * 根据俩个时间获取俩个时间的 包含的  年,月数,天数,小时,分钟,秒
	 * @param String $start
	 * @param String $end
	 * @return ArrayObject 
	 */
	 private function diffDateTime($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

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