這應該適合你。這將爲每個學生分配一個獨特的present值，然後在回發時對其進行檢查，如果設置，則將其清理並用於更新出席的學生記錄。
我還將PHP中的echo'd HTML解壓縮爲HTML，並將表單移動到表格外（這可能會導致某些瀏覽器出現問題）。
<?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=" .$rows['present'] . ">";
}
?>
</table>
<input type='submit' name='Submit' value='Submit'>
</form>
替代（更好的）方法：如果本值可以設置爲一個簡單的0/1或真/假，那麼這將是更容易使用複選框爲每個學生。在回發中，您可以通過檢查每個指示存在的學生的複選框來檢索一組值，然後在一個查詢中更新數據庫表。這也可以防止惡意文本輸入。
替代代碼：
<?php
// Update present values
if (isset($_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>
你連接到數據庫？ – Vucko 2013-03-12 20:01:17
是的，我連接到數據庫 – 2013-03-12 20:03:02
它看起來像你正在更新所有學生的現值，因爲你唯一的WHERE檢查是'course_id'和'week_id'。如果while循環包含多條記錄，那麼您將擁有元素' 2013-03-12 20:03:53