这应该适合你。这将为每个学生分配一个独特的present值，然后在回发时对其进行检查，如果设置，则将其清理并用于更新出席的学生记录。
我还将PHP中的echo'd HTML解压缩为HTML，并将表单移动到表格外（这可能会导致某些浏览器出现问题）。
<?php
// Update present values
if (isset($_POST['submit']))
{
// Get a list of student ids to check
$idsResult = mysql_query("SELECT student_id from students");
while($idRow = mysql_fetch_array($idsResult))
{
// if the textbox for this student is set
if(isset($_POST['present'.$idRow['student_id']]) && !empty($_POST['present'.$idRow['student_id']]))
{
// Clean the user input, then escape and update the database
$cleanedPresent = htmlspecialchars(strip_tags($_POST['present'.$idRow['student_id']]));
$sql = "UPDATE course_attendance SET present='".mysql_real_escape_string($present)."' WHERE course_id='101' AND week_id='2' AND student_id=".$idRow['student_id'];
$result = mysql_query($sql);
if($result){
echo "Successfully logged the attendance for ID ".$idRow['student_id'];
}
else {
echo "ERROR updating on ID ".$idRow['student_id'];
}
}
}
}
$test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_attendance.week_id=attendance.week_number_id AND course_attendance.student_id= students.student_id AND courses.course_id='101' AND attendance.week_number_id='2' ";
$result = mysql_query($test3) or die(mysql_error());
?>
<form name='Biology_lecture11.php' method='post'>
</br>
<table border='1' align='center'>
<tr>
<th><strong>Student ID</strong></th>
<th><strong>First Name </strong></th>
<th><strong>Last Name</strong></th>
<th><strong>Present</strong></th>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
echo "<tr><td width='100' align='center'>" .$rows['student_id'].
"</td><td width='120' align='center'>" .$rows['fname'].
"</td><td width='120' align='center'>" .$rows['lname'].
"</td><td><input type='text' name='present".$rows['student_id']."' value=" .$rows['present'] . ">";
}
?>
</table>
<input type='submit' name='Submit' value='Submit'>
</form>
替代（更好的）方法：如果本值可以设置为一个简单的0/1或真/假，那么这将是更容易使用复选框为每个学生。在回发中，您可以通过检查每个指示存在的学生的复选框来检索一组值，然后在一个查询中更新数据库表。这也可以防止恶意文本输入。
替代代码：
<?php
// Update present values
if (isset($_POST['submit']))
{
// Get a list of student ids to check
$idsResult = mysql_query("SELECT student_id from students");
$presentIds = array();
$absentIds = array();
while($idRow = mysql_fetch_array($idsResult))
{
// If the student's checkbox is checked, add it to the presentIds array.
if(isset($_POST['present'.$idRow['student_id']]))
{
$presentIds[] = $idRow['student_id'];
}
else
{
$absentIds[] = $idRow['student_id'];
}
}
// Convert array to string for query
$idsAsString = implode(",", $presentIds);
// You can set present to whatever you want. I used 1.
$sql = "UPDATE course_attendance SET present='1' WHERE course_id='101' AND week_id='2' AND student_id IN (".$idsAsString.")";
$result = mysql_query($sql);
if($result){
echo "Successfully logged the attendance for IDs ".$idsAsString;
}
else {
echo "ERROR updating on IDs ".$idsAsString;
}
// OPTIONAL: Mark absent students as '0' or whatever other value you want
$absentIdsAsString = implode(",", $absentIds);
// You can set present to whatever you want. I used 1.
$absentQuery = "UPDATE course_attendance SET present='0' WHERE course_id='101' AND week_id='2' AND student_id IN (".$absentIdsAsString.")";
$absentResult = mysql_query($absentQuery);
if($absentResult){
echo "Successfully logged absence for IDs ".$absentIdsAsString;
}
else {
echo "ERROR updating absence on IDs ".$absentIdsAsString;
}
}
$test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_attendance.week_id=attendance.week_number_id AND course_attendance.student_id= students.student_id AND courses.course_id='101' AND attendance.week_number_id='2' ";
$result = mysql_query($test3) or die(mysql_error());
?>
<form name='Biology_lecture11.php' method='post'>
</br>
<table border='1' align='center'>
<tr>
<th><strong>Student ID</strong></th>
<th><strong>First Name </strong></th>
<th><strong>Last Name</strong></th>
<th><strong>Present</strong></th>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
echo "<tr><td width='100' align='center'>" .$rows['student_id'].
"</td><td width='120' align='center'>" .$rows['fname'].
"</td><td width='120' align='center'>" .$rows['lname'].
"</td><td><input type='checkbox' name='present".$rows['student_id']."' ";
// NOTE: REPLACE 1 with whatever value you store in the database for being present.
// I used 1 since the update at the top of the code uses 0 and 1.
if($rows['present']=='1')
{
echo "checked='checked' ";
}
// With a checkbox, you don't need to assign it a value.
echo "value=" .$rows['present'];
echo ">";
}
?>
</table>
<input type='submit' name='Submit' value='Submit'>
</form>
你连接到数据库？ – Vucko 2013-03-12 20:01:17
是的，我连接到数据库 – 2013-03-12 20:03:02
它看起来像你正在更新所有学生的现值，因为你唯一的WHERE检查是'course_id'和'week_id'。如果while循环包含多条记录，那么您将拥有元素' 2013-03-12 20:03:53