これは、特定の日付範囲で従業員が取得した休暇のリストを取得するために作成した関数です。取ったリーフが1枚でも2枚でもいいのですが、複雑すぎて結果の取得に時間がかかってタイムアウトエラーになってしまいます！何か助けはありますか？
これは機能です：
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);
}