Question

我想总结日期差异，一切都很好，但是如果有相同的日期我想加1，例如，如果起始日期是：01/01/2003 并且到目前为止是01/01/2003然后我想要添加1天，但它不会增加1天，而是只增加一天，如果是01/01/03到01/02/2003。这是计算休假，如果有人申请休假一天，即2003年1月1日至2003年1月1日，它应计算1天

＆＃13;

	$sql = "SELECT l.id as typeid,l.*,ud.firstname, ud.lastname,ud.email,l.leave_starting_from,l.leave_partial_from,ud.joining_date, l.typename, l.no_of_leave_days,(CASE SUM(datediff(STR_TO_DATE(lr.to_date, '%m/%d/%Y'), STR_TO_DATE(lr.from_date, '%m/%d/%Y'))) when NULL then '0' else SUM(datediff(STR_TO_DATE(lr.to_date, '%m/%d/%Y'), STR_TO_DATE(lr.from_date, '%m/%d/%Y'))) end) as days_in_betw,  lr.from_date, lr.to_date FROM pr_users u INNER JOIN pr_users_details ud on ud.userid= u.id LEFT OUTER JOIN pr_leave_type l ON ((FIND_IN_SET( u.departmentid , l.function ) >0) AND (FIND_IN_SET( ud.designation_id , l.level ) >0) and (l.gender='All' or l.gender=ud.gender)) LEFT OUTER JOIN pr_leave_request lr ON u.id = lr.userid and lr.sr_type = l.id";

＆＃13;

特别是其他部分

＆＃13;

SUM(datediff(STR_TO_DATE(lr.to_date, '%m/%d/%Y'), STR_TO_DATE(lr.from_date, '%m/%d/%Y'))

＆＃13;

Answer 1

我遇到过同样的问题，愿这个功能解决你的问题。

function days_calculation($start_date,$end_date){
  $start = new DateTime($start_date);
  $end = new DateTime($end_date);
  // otherwise the  end date is excluded (bug?)
  $end->modify('+1 day');

  $interval = $end->diff($start);

  // total days
  $days = $interval->days;

  // create an iterateable period of date (P1D equates to 1 day)
  $period = new DatePeriod($start, new DateInterval('P1D'), $end);

  // best stored as array, so you can add more than one
  $holidays = array();

  foreach($period as $dt) {
      $curr = $dt->format('D');

      // substract if Saturday or Sunday
      if ($curr == 'Sat' || $curr == 'Sun') {
          $days--;
      }

      // (optional) for the updated question
      elseif (in_array($dt->format('Y-m-d'), $holidays)) {
          $days--;
      }
  }

  if($days==1){
      $days = $days." day";
  } else {
      $days = $days." days";

  }

  return $days;
}

Answer 2

我认为这可能会有所帮助。它在T-SQL中，因此可能需要重新调整。

/* Setting up your problem */
create table #temp
(
personID int,
from_date datetime,
to_date datetime
)

insert into #temp values (1,'01/01/2003','01/01/2003')
insert into #temp values (1,'01/01/2013','05/01/2013')
insert into #temp values (2,'01/01/2013','05/01/2013')


/* Here the nested statement calculates the number of days off, then the outer select group this per personID */
select  personID
    ,   SUM(DaysOff) as TotalDaysOff
from    (
        select  personID
            ,   datediff(day, lr.from_date, lr.to_date)+1 as DaysOff
        from    #temp lr
        ) lr
group by personID

您需要将其更改为MySQL，然后将其放在您拥有的select语句中。

具有相同日期的Datediff总和

2 个答案: