核心代码:
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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