Question

我的代码计算当月缺勤的工作日如果在月份我需要计算缺勤只有一个人的期间来自和将来的日期保存在数据库中示例2016-08-09 - 2016-08-13（假设用户病了）它计算4个工作日

但如果用户有：

2016-07-27 - 2016-08-02（假设用户病了）
2016-08-08 - 2016-08-10（假设用户在职业）和
2016-08-24 - 2016-09-02（再说让用户生病了）
如何统计2016年8月的缺席期间的所有日子？在这种情况下应该是12天

从数据库获取：

 $getaways =  mysqli_query($conn, "SELECT * FROM employeesAbsence WHERE workerID='".$row['worker']."' AND fromDate LIKE  '%2016-08%' AND toDate LIKE  '%2016-08%'); 
 $aways_row = mysqli_fetch_array($getaways);

        $fDate = $aways_row['fromDate'];
        $tDate = $aways_row['toDate'];

计算代码：

$startDate = $fDate;
$endDate =  $tDate; 
$weekdays = array('1','2','3','4','5'); //monday-friday

$begin = new DateTime($startDate);
$end = new DateTime($endDate);

$end = $end->modify( '+1 day' ); //add one day so as to include the end date of our range

$interval = new DateInterval('P1D'); // 1 Day
$dateRange = new DatePeriod($begin, $interval, $end);

$total_days = 0;
//this will calculate total days from monday to friday in above date range
foreach ($dateRange as $dates) {

 if (in_array($dates->format("N"),$weekdays)) {
        $total_days++; 
  }
}
echo $total_days;

Answer 1

您可能完全在sql查询中执行此操作 - 未经测试但我认为应该没问题。

例如：

select sum( datediff(`toDate`,`fromDate`) ) as 'sickdays' 
   from `employeesabsence`
   where `workerID`=$row['worker']

您可以在where子句中添加其他条件以限制特定日期范围 - 例如：

select sum( datediff(`todate`,`fromdate`) ) as 'sickdays' 
    from `employeesabsence` 
    where `workerid`=$row['worker'] and month( `fromdate` ) between 1 and 6;

在sql中找到工作日的解决方案有点棘手，以下内容改编自堆栈中发现的另一个帖子 - 它涉及一个奇特的整数字符串。

例如

select 
sum( 5 * ( datediff( `todate`, `fromdate` ) div 7 ) + mid( '0123455401234434012332340122123401101234000123450', 7 * weekday(`fromdate`) + weekday(`todate`) + 1, 1) ) as 'sickdays'
from `employeesabsence` where month(`fromdate`) between 6 and 7

对其进行调整的帖子可以是found here

Answer 2

PHP DateTime选项：

    function get_days( $start, $end )
    {  
      $begin = new DateTime( $start );
      $end   = new DateTime( $end );
      //$end   = $end->modify( '+1 day' ); //add one day so as to include the end date of our range

      $total_days = 0;
    //this will calculate total days from monday to friday in above date range
      for( $i = $begin; $i <= $end; $i->modify( '+1 day' ) )
      {
        // Check that the date is between Monday and Friday and only in August
        if( ( $i->format( 'N' ) >= 1 && $i->format( 'N') <= 5 ) && $i->format( 'm' ) == '08' )
        {
          $total_days++;
        }
      }
      return $total_days;
    }

    $total = 0;
    $total += get_days( '2016-07-27', '2016-08-02');
    $total += get_days( '2016-08-08', '2016-08-10');
    $total += get_days( '2016-08-24', '2016-09-02');

在一个月内计算多个时段/缺勤天数

2 个答案: