PHP & MYSQL Return blank page - php

Question

I'm working on the PHP and MYSQL for a project and I encountered a weird problem here where I click on the submit button at the form, it will run these codes. However the weird problem is the the page return blank instead of going back to the page with the form. I had searched for few hours for the error but couldn't find it.
Please point out my mistake. Thank you for the help.
<?php
include '../database.php';
if(isset($_POST['submit'])) {
if (isset($_POST['stuid_0'])){
$student = $_POST['stuid_0'];
//query moderator details
$query = mysql_query(" SELECT ModeratorID FROM Student WHERE StuID ='$student' ") or die(mysql_error());
$info = mysql_fetch_assoc ($query);
$dbmoderator = $info['ModeratorID'];
//check for changes of status in supervisor
$query2 = mysql_query(" SELECT SupervisorID FROM Student WHERE StuID ='$student' ") or die(mysql_error());
$value = mysql_fetch_assoc ($query2);
$dbsupervisor = $value['SupervisorID'];
$query3 = mysql_query(" SELECT LectStatus FROM Lecturer WHERE LectID ='$dbsupervisor' ") or die(mysql_error());
$value2 = mysql_fetch_assoc ($query3);
$dbsupervisorstatus = $value2['LectStatus'];
//if no changes in supervisor
if ($dbsupervisorstatus=='2'){
echo ("<SCRIPT LANGUAGE='JavaScript'>
window.alert('Moderator can't be promoted')
window.location.href='../committee/committee_supervisor2.php'
</SCRIPT>");
}
else{
//newly assigned a supervisor if previous supervisor status is not active
$query4 = "UPDATE Student SET SupervisorID='$dbmoderator', SupervisorStatus='1', ModeratorID=NULL WHERE StuID='$student'";
mysql_query($query4);
echo ("<SCRIPT LANGUAGE='JavaScript'>
window.alert('Successfully updated')
window.location.href='../committee/committee_supervisor2.php'
</SCRIPT>");
}
}
else
echo ("<SCRIPT LANGUAGE='JavaScript'>
window.alert('You must choose a moderator to be promoted')
window.location.href='../committee/committee_supervisor2.php'
</SCRIPT>");
}
?>
UPDATE:
I think the problem happen when the system run at this point
if ($dbsupervisorstatus=='2'){
As i put an echo "Test"; before this line and it still work.
UPDATE 2:
I found that the code can be run when I put
if ($dbsupervisorstatus=='2'){
echo "Moderator can't be promoted";
}
as well as
if($dbsupervisorstatus == 2){
header("location:commitee_supervisor2.php");
}
However I don't see the reason why my original code
if ($dbsupervisorstatus=='2'){
echo ("<SCRIPT LANGUAGE='JavaScript'>
window.alert('Moderator can't be promoted')
window.location.href='../committee/committee_supervisor2.php'
</SCRIPT>");
}
Not working.. A little help pls.. :)
FINAL UPDATE
Guys, I know why.
It's because the
window.alert('Moderator can't be promoted')
have 3 apostrophe in it.
I simply remove the word "can't" and its working already.
Thank you guys for the help :)

Answer 1

Using the JavaScript codes code still work but the starting point of your debugging is to echo ordinary texts at all the places where you echo JavaScript. That will help you know at what point you codes started failing. Example
//if no changes in supervisor
if ($dbsupervisorstatus=='2'){
echo ("<SCRIPT LANGUAGE='JavaScript'>
window.alert('Moderator can't be promoted')
window.location.href='../committee/committee_supervisor2.php'
</SCRIPT>");
}
Replace with
//if no changes in supervisor
if ($dbsupervisorstatus=='2'){
echo "Moderator can't be promoted";
}
Do this at all echo points then start replacing one after the other with your JavaScript codes again then you'll spot out where your code fails.

Answer 2

Instead of doing it through javascript, try using a much better approach in pure PHP. Something like:
if($dbsupervisorstatus == 2){
header("location:commitee_supervisor2.php");
}
And if you must use JS, then make an AJAX request, then you can manipulate the behaviour through JS anyway you like.

Answer 3

Where do you put your HTML form? Is it in the same page or different page? From your code, looks like there's nothing being done, so I suggest you to check whether your submit button name is correct (submit). The HTML should be like this:
<form name="your_form">
<!-- Your form here -->
<input type="submit" name="submit" value="Submit"/>
</form>