所以我有一些代码,似乎我必须一遍又一遍地重复 $round == P,1,2 和 A。有没有我可以在不产生冗余代码的情况下实现这一点?
我正在考虑编写一个函数,它可以简单地交换所有 3 个可用的 $round 变量并回显相关信息。关于如何实现这一目标的任何想法?
其他信息将保持不变 - 它唯一的 $round 会改变,
// determine previous round
if($round == "P") echo "";
// if we are in round 1, look up round P bookings
if($round == "1")
{
$sql = "SELECT
*
FROM ts_request
INNER JOIN ts_day
ON ts_request.day_id = ts_day.id
INNER JOIN ts_period
ON ts_request.period_id = ts_period.id
INNER JOIN ts_allocation
ON ts_request.id = ts_allocation.request_id
WHERE ts_request.round=:round
AND ts_request.dept_id=:dept
ORDER BY ts_request.module_id ASC";
$stm = $pdo->prepare( $sql );
$stm->execute( array( ':round' => 'P', ':dept' => $loggedin_id ) );
$rows = $stm->fetchAll();
foreach ($rows as $row)
{
echo '<tr align="center">';
echo '<td>'.$row['module_id'].'</td>';
echo '<td>'.$row['day'].'</td>';
echo '<td>'.$row['period'].'</td>';
echo '<td>';
$sql = "SELECT * FROM ts_roompref
WHERE request_id=:id";
$stm = $pdo->prepare( $sql );
$stm->execute( array( ':id' => $row['request_id']) );
$rows2 = $stm->fetchAll();
foreach ($rows2 as $row2)
{
if ($row2['room_id']=="0")
{
echo "Any<br>";
}
else
{
echo $row2['room_id'].'<br>';
}
}
echo '</td>';
echo '<td>'.$row['status'].'</td>';
echo '</tr>';
}
}
// if we are in round 2, look up round 1 bookings
if($round == "2")
{
$sql .= "";
}
foreach ($rows as $row)
{
// echo results here
};
// if we are in round A, look up round 2 bookings
if($round == "A")
{
$sql .= "";
}
foreach ($rows as $row)
{
// echo results here
};