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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 ;
}
}
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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