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这是我为获取员工在给定日期范围内的休假日列表而创建的功能。如果取的叶子是一两片也没关系，但它太复杂了，检索结果需要很长时间，因此会导致超时错误！有什么帮助吗？

这是功能：

function dates_between($emp_id, $start_date, $end_date) 
 {

    $day_incrementer = 1;
    $count_leaves = 0;
    $flag = 0;

    // Getting the days from DB where the employee '28' had worked in given date range

    $work_res = mysql_query("SELECT DISTINCT date FROM `work_details` WHERE  employee_id='28' and date between '2012-02-01' and '2012-02-29'");

    do {
        while($row = mysql_fetch_array($work_res)) 
             {
           while((date("Y-m-d",$start_date) < $row['date']) && ($flag = 0)) 
                       // loop to find  startdate  less than table date! if table date(attendance) is starting from 3, we need to print leaves 1,2  if they are not  weekends
                   {
                 if(!(date('N', strtotime(date("Y-m-d", $start_date))) >=6)) 
                       {    
                               //checking for weekends, prints only weekdays
                    echo date("Y-m-d", $start_date) . " \n ";
                    $count_leaves++;
               }

           $start_date = $start_date + ($day_incrementer * 60 * 60 *24);              
            }

            $flag=1;


    while((date("Y-m-d",$start_date) != $row['date']))
     // loop to print $start_date,which is not equal to table date
    {
    if(!(date('N', strtotime(date("Y-m-d", $start_date))) >= 6)) 
      {
        echo  date("Y-m-d", $start_date) . "\n";
        $count_leaves++;
      }
     $$start_date = $start_date + ($day_incrementer * 60 * 60 * 24);
     }

        $start_date = $start_date + ($day_incrementer * 60 * 60 * 24);
    }

 // loop to print $start_date,comes rest after tabledate if tabledate finishes with 28, prints rest of dates 29,30
  if(!(date('N', strtotime(date("Y-m-d", $start_date))) >= 6) && ($start_date <= $end_date))
  {
            echo  date("Y-m-d", $start_date) . "\n";
            $count_leaves++;
            $start_date = $start_date + ($day_incrementer * 60 * 60 * 24);
  }


  } while($start_date <= $end_date);

    return($count_leaves);
 }
4

1 回答 1

0

我注意到您在其他地方也问过类似的问题（http://stackoverflow.com/questions/10898293/how-to-get-days-of-leave-taken-in-a-given-month）。现在，我尝试深入研究您的代码，以基本了解您的尝试。如果我的回答不完全符合您的愿望，请原谅我，因为不容易读懂另一个人的想法。基本上，我所做的是准备一个示例代码来满足您的需求。此代码采用特定工作人员在给定月份和年份中工作的日期数组。然后它会继续获取该给定月份、那一年的所有可用工作日期。两个数组的差异给出了工人缺席的日期（由于离开或擅离职守）。没有考虑公共假期，但当然，您可以轻松修改代码以添加它。

现在，只是一个警告，这段代码是基本的，如果两个数组的日期格式不完全相同，数组差异会让你失败。就个人而言，我将编写自己的比较回调函数来比较各个日期并将其传递给 array_udiff() 以获得最大的确定性。我很确定你能处理好。我只提供了基础知识。自由使用并根据您的情况进行扩展。说得够多了，请看下面的代码示例。

<?php
/***************************************************************************
* how to get DAYS absent from working days from given date range?
* @Author Prof. No Time - 12th/June/2012
****************************************************************************/

//Leave was 10th, 13th, 23rd, 24th
//Note that 01-02-2012 is NOT exactly same as 1-2-2012; Important for the array_diff fxn used below. 
//Note Format is d-m-Y
//Note I am assuming you have pulled this from a database of course
$imaginaryWorkDatesOfWorker1 = array(
    '01-02-2012', '02-02-2012', '03-02-2012', '06-02-2012', '07-02-2012', '08-02-2012',
    '09-02-2012', '14-02-2012', '15-02-2012', '16-02-2012', '17-02-2012', '20-02-2012',
    '21-02-2012', '22-02-2012', '27-02-2012', '28-02-2012', '29-02-2012'    
);

$leaveDays1 = getLeaveDays(2, 2012, $imaginaryWorkDatesOfWorker1);
displayWorkersLeaveDays($leaveDays1);

//Leave was 2nd, 16th, 19th, 23rd and 26th
$imaginaryWorkDatesOfWorker2 = array(
    '01-03-2012', '05-03-2012', '06-03-2012', '07-03-2012', '08-03-2012', '09-03-2012',
    '12-03-2012', '13-03-2012', '14-03-2012', '15-03-2012', '20-03-2012', '21-03-2012',
    '22-03-2012', '27-03-2012', '28-03-2012', '29-03-2012', '30-03-2012'
);

$leaveDays2 = getLeaveDays(3, 2012, $imaginaryWorkDatesOfWorker2);
displayWorkersLeaveDays($leaveDays2);



///MAIN FUNCTION TO GET LEAVE DATES///
function getLeaveDays($month, $year, $arrDatesPresent=array()){
  $arrAllWorkDatesInMonth = getDatesInTheMonth($month, $year);

  //var_dump($arrDatesPresent); var_dump($arrAllWorkDatesInMonth);

  $leaveDays = array_diff($arrAllWorkDatesInMonth, $arrDatesPresent);
  return $leaveDays;
}


///HELPER FUNCTIONS///
/**
 * <p>Gets all the dates in a given month in the specified year. default format d-m-Y<p>
 * @param int $month
 * @param int $year
 * @param boolean $includeWeekends
 * @param string $format2Use
 * @throws Exception if invalid parameters are given
 * @return array: dates in the given month, in the given year
 */
function getDatesInTheMonth($month, $year, $includeWeekends=false, $format2Use='d-m-Y')    {
  $arrDatesInTheMonth = array();
  if (empty($format2Use)) $format2Use = 'm-d-Y';

  if (empty($month) || empty($year)){
    throw new Exception("Invalid parameters given.");
  }
  else{
    $fauxDate = mktime(0, 0, 0, $month, 1, $year);
    $numOfDaysInMonth = date('t', $fauxDate);

    if (!empty($numOfDaysInMonth)){
        for ($day = 1; $day <= $numOfDaysInMonth; $day++){

            $timeStamp = mktime(0, 0, 0, $month, $day, $year);
            $cdate = date($format2Use, $timeStamp);

            if ($includeWeekends){
                $arrDatesInTheMonth[] = $cdate;
            }
            else{
                if (!isWeekend($cdate)) { $arrDatesInTheMonth[] = $cdate; }
            }
        }
    }
  }

  return $arrDatesInTheMonth;
}

/**
 * Checks if given date is a weekend use this if you have PHP greater than v5.1.
 * Credit: http://stackoverflow.com/users/298479/thiefmaster
 * @param date $date
 * @return boolean
 */
function isWeekend($date) {
  return (date('N', strtotime($date)) >= 6);
}


/**
 * Checks if given date is a weekend use this if you have PHP less than v5.1.
 * Credit: http://stackoverflow.com/users/298479/thiefmaster
 * @param date $date
 * @return boolean
 */
function isWeekend2($date) {
  $weekDay = date('w', strtotime($date));
  return ($weekDay == 0 || $weekDay == 6);
}

function printDates($arrDates){
  foreach ($arrDates as $key => $cdate) {
      $display = sprintf( '%s <br />', date('[l] - jS \of F Y', strtotime($cdate)) );
      echo $display;
  }
}

function displayWorkersLeaveDays($leaveDays){
  echo '<div style="background-color:#CCC;margin:10px 0;">';
  echo '<div>Your Leave days are as follows: </div>';
  printDates($leaveDays);
  echo '</div>';
}

希望这可以帮助。

于 2012-06-12T18:32:59.407 回答