如何从foreach循环中的数据库中获取日期值

时间:2017-09-06 02:15:24

标签: php mysql

它已经工作但仅限假期。我已将foreach循环中的数据库连接起来并减去罚款the table in my database中的假期,我网站中的表已将其视为假期table in website但在我添加另一个之后在9月6日的数据库中度假,结果就像这个“imgur.com/a/l8Deq”不算假期,不知道为什么我的罚款和日子都在增加。我该怎么办?谢谢你的帮助

$borrowdate = new Datetime($row['date_return'],new DateTimeZone('Asia/Manila'));
$returndate = new Datetime($row['due_date'],new DateTimeZone('Asia/Manila')); 
$currentdate = new Datetime('Asia/Manila');

$returndate->setTime(0,0);
$currentdate->setTime(0,0);
$borrowdate->setTime(0,0);

$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($returndate, $interval, $borrowdate);

$borrowdate->format("D");
$weekendDays = 0;
$totalDays = 0;
$holiDays = 0;


$query_holiday =mysqli_query($dbcon,"SELECT * FROM holiday_tbl"); 
while($row=mysqli_fetch_array($query_holiday)){                   
$holi = $row ['holiday'];                                          

foreach ($period as $p ) 
{
   $totalDays++;
   if($p->format( "w" )== 0 or $p->format( "w" )==6 ) $weekendDays++;
   if($p->format('Y-m-d') == $holi) $holiDays++;

 }
 }
echo "<p>Total  days: $totalDays</p><p>Weekend days: $weekendDays </p> <p>                 Holidays: $holiDays </p>";
$fines = ($totalDays - $weekendDays - $holiDays) * 5;
echo "₱ ". $fines;
$fi = $row['borrow_details_id'];
mysqli_query($dbcon,"update borrowdetails set fines='$fines' where borrow_details_id = '$fi'");

0 个答案:

没有答案