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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