PHP gets the year, month, day, hour, minute, and second in the course of the specified time period

PHP gets the year, month, day, hour, minute, and second between the specified time period
Requirement: The front end passes two standard time formats, formatted like 2009-05-12 12:12:30, then returns the representation of different units of this time period as needed
For the time format of the verification I did not post the code, so use when you consider adding

Class Utils {/** * format mysql DateTime (yyyy-mm-dd hh:mm:ss) converts the data format found in MySQL into time seconds * @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);} /** * * * * * * * * * * * * * * * * * for two time inclusive years, months, days, hours, minutes, seconds * @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;//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;//year if ($eMonth-$sMonth <0) {//month minus 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 an example, 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 possible

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