How PHP gets the year, month, day, hour, minute, and second code instance between the specified time period

The front end passes two standard time format come over, format like 2009-05-12 12:12:30, then return according to need different units of this time period of representation, for time format of check I here code is not posted out, so use time oneself consider add

Core code:

Class Utils {/** * format mysql DateTime (yyyy-mm-dd hh:mm:ss) converts the data format found in MySQL into time seconds * @param string $datetime */pu  Blic function Fmdatetime ($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);} /** * * * * * * * * * * * * * * * * * for two time inclusive years, months, days, hours, minutes, seconds * @param string $start * @param string $end * @return Arrayobject */Private F Unction diffdatetime ($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;//Get two time seconds $d = Ceil ($dis/(24*60*60));//Get Days $rs[' Day ' = $d;//Days $rs[' hour '] = Ceil ($dis/(60*60)) ;//Hours $rs[' minute ']= Ceil ($dis/60);//min $rs[' second '] = $dis;//Seconds $rs[' Week '] = ceil ($d/7);//week $tem = ($eYear-$sYear) *12;//Month $tem1 = $eYear-$ syear;//if ($eMonth-$sMonth <0) {//month subtracts negative $tem + = ($eMonth-$sMonth);} else if ($eMonth = = $sMonth) {//month same if ($eDay-$sDay >=0) {$tem + +; $tem 1++;}} else if ($eMonth-$sMonth >0) {//month minus $tem1++;if ($eDay-$sDay >=0) {//and date subtracted to positive $tem + = ($eMonth-$sMonth) +1;} else{$tem + = ($eMonth-$sMonth);}} $rs [' month '] = $tem; $rs [' year '] = $tem 1;return $rs;}}

More than a year, the return is 2 years, one months more than a day to return is 2 months, to push ... Project needs, just do this out, I also go to the Internet to find such examples, but everyone is the year on the basis of 365 days to calculate, the month according to 30 days to calculate, so that the results are certainly useless, the year may be 366 days, the month may be 31,29,28 are likely to