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['month'] = $tem; $rs['year'] = $tem1; return $rs; } }
One more day in a year will return 2 years, and one more day in a month will return 2 months. Based on this... I only made this when the project needed it. At first, I also looked for something like this on the Internet. Example, but everyone calculates the year as 365 days, and the month as 30 days. The result of this calculation is definitely useless. The year may be 366 days, and the month may be 31, 29, or 28. It’s possible