我需要从数据库中检索相关数据，然后再次添加到考勤表 . 此代码显示许多错误 . 实际上我不知道如何使用数组添加数据 . 谁能帮忙？ thankssssssssssss .
<?php
$sql = "select p_module_ID, student_ID,
p_attendance_Date, p_attendance_Time,
p_attendance_Status,p_attendance_reason
from attendance";
$result = mysqli_query($con, $sql);
if(!$result)
{
echo mysqli_error($con);
exit();
}
while($rows = mysqli_fetch_array($result))
{
$attendance_list[] = array('p_module_ID' => $rows['p_module_ID'],
'student_ID' => $rows['student_ID'],
'p_attendance_Date' => $rows['p_attendance_Date'],
'p_attendance_Time' => $rows['p_attendance_Time'],
'p_attendance_Status' => $rows['p_attendance_Status'],
'p_attendance_reason' => $rows['p_attendance_reason']);
}
?>
<html>
<body>
<form action="attendance.php" method="post" accept-charset='UTF-8'>
<table border="0" cellspacing="20" >
<tr>
<td>
<select name="p_module_ID">
<?php
$sql = "SELECT p_module_ID FROM schedule";
$result = $con->query($sql);
while($row = mysqli_fetch_assoc($result)) {
echo "<option>".$row['p_module_ID']."</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td>
<select name="p_attendance_Date" >
<?php
$sql = "SELECT p_StartDate FROM schedule";
$result = $con->query($sql);
while($row = mysqli_fetch_assoc($result)) {
echo "<option>".$row['p_StartDate']."</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td>
<select name="p_attendance_Time">
<?php
$sql = "SELECT p_Time FROM schedule";
$result = $con->query($sql);
while($row = mysqli_fetch_assoc($result)) {
echo "<option>".$row['p_Time']."</option>";
}
?>
</select>
</td>
</tr>
<table id="t01">
<tr>
<th> Student ID </th>
<th> Name </th>
<th> Attendance </th>
<th> Reason </th>
<?php foreach($attendance_list as $attend) : ?>
<tr>
<td>
<?php
$sql = "SELECT p.student_ID,CONCAT(s.student_fname, ' ', s.student_lname) AS fullname FROM Pals p JOIN student s ON p.student_ID = s.student_ID WHERE student_role = 'Student' GROUP BY student_ID";
$result = $con->query($sql);
while($row = mysqli_fetch_assoc($result)) {
echo $attend[$row['student_id']];
}
?>
</td>
<td>
<?php
$sql = "SELECT p.student_ID,CONCAT(s.student_fname, ' ', s.student_lname) AS Fullname FROM Pals p JOIN student s ON p.student_ID = s.student_ID WHERE student_role = 'Student' GROUP BY student_ID";
$result = $con->query($sql);
while($row = mysqli_fetch_assoc($result)) {
echo $attend['Fullname'];
}
?>
</td>
<td>
<?php echo $attend["p_attendance_Status"]; ?>
</td>
<td>
<?php echo $attend["p_attendance_reason"]; ?>
</td>
</tr>
</tr>
<?php endforeach; ?>
</table>
</table>
</form>
</body>
</html>
1 回答
使用
$row['student_Id']而不是$row['student_id']. 与Fullname相同 . 您选择AS fullname而不是AS Fullname. 听取案例敏感性 .