Home Backend Development PHP Tutorial How to get the year, month, day, hour, minute and second code example between a specified time period in PHP

How to get the year, month, day, hour, minute and second code example between a specified time period in PHP

Jul 10, 2017 am 09:45 AM
php period

The front end passes two standard time formats, the format is like 2009-05-12 12:12:30, and then returns the representation of different units of this time period as needed. I have not posted the code here for the verification of the time format. come out, so when using it, consider adding

core code:

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[&#39;month&#39;] = $tem;
		$rs[&#39;year&#39;] = $tem1;
		
		return $rs;
	}
}
Copy after login

One more day in a year returns 2 years, and one more day in a month returns 2 months. I did this because of project needs. At first, I looked for such examples on the Internet, but everyone calculated the year as 365 days and the month as 30 days. This is how it was calculated. The result is definitely useless. The year may have 366 days, the month may have 31, 29, or 28.

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