mysql database not updating

[raident30]

hi guys! i have these codes.. the select from database is working but it wont update.. any chance that im doing wrong?
here are the codes


configdb.php

<?php
$hostname ="localhost";
$username ="root";
$password ="";
$db ="enrollmentdb";
mysql_connect($hostname,$username,$password) or die("cannot connect to server");
mysql_select_db($db) or die("database not found!");
?>

prelimgrading.php

<?php




	
include "configdb.php";
$sql = "SELECT enrollment.student_num, students.lastname, students.firstname, students.middlename, enrollment.subject_num, prelim
		FROM enrollment
		LEFT OUTER JOIN students ON enrollment.student_num = students.student_num;";
$result = mysql_query($sql) or die ("cannot execute query!");
	$count = 0;
	$vline = 1;
	echo "<form action = 'updateprelimgrade.php' method = 'POST' onsubmit = 'return checkform(this)'; >";
	while($row = mysql_fetch_row($result)){
		$student_num = $row[0];
		$lastname = $row[1];
		$firstname = $row[2];
		$middlename = $row[3];
		$subject_num = $row[4];
		$prelim = $row[5];
		
		
	
		
		$count = $count+1;
		
		if($vline%2==1) {
			$bg_color = "#3BB9FF";
			$vline = 2; }
			
					else
					{
						$bg_color="#2554C7";
						$vline=1;
						}
		
		
		

		print "<div class='tablecontents' style='float:left';>";
		print "<table align = 'left' height = '25' width='750' 'fontko'>";
		print "<center>";
		print "<tr>";
		echo "<td width = '100' bgcolor = '$bg_color'><p align = 'left'><font size = '2' face = 'Century Gothic' color ='white'> <input type='text' id='student_num' name='student_num' value=$student_num></font></td> </font></td>";
		print "<td width = '100' bgcolor = '$bg_color'><p align = 'left'><font size = '2' face = 'Century Gothic' color ='white'> $lastname </font></td>";
		print "<td width = '100' bgcolor = '$bg_color'><p align = 'left'><font size = '2' face = 'Century Gothic' color ='white'> $firstname </font></td>";
		print "<td width = '100' bgcolor = '$bg_color'><p align = 'left'><font size = '2' face = 'Century Gothic' color ='white'> $middlename </font></td>";
		echo "<td width = '150' bgcolor = '$bg_color'><p align = 'left'><font size = '2' face = 'Century Gothic' color ='white'><input type='text' id='prelim' name='prelim' value=$prelim></font></td>";
			
		print"</tr>";
		print "</center>";
		print"</table>";
		print"</div>";
		}
	print "<input type='submit' value='Submit'>";
	echo "</form>";
						
	?>

updateprelimgrade.php

<?php
$con = mysql_connect('localhost','root','') or die('Unable to select database');

mysql_select_db("enrollmentdb", $con);
$student_num=$_POST['student_num'];
$prelim=$_POST['prelim'];
$query="update enrollment SET prelim=$prelim where student_num=$student_num";
$result = mysql_query($query) or die ("Cannot Execute Query!");
if (!mysql_query($result))
{
echo"Failed to Update the Grading";

}
else
header('Location: prelimgrading.php');

?>

these are the database tables from the "enrollmentdb" database

CREATE TABLE IF NOT EXISTS `enrollment` (
  `STUDENT_NUM` varchar(10) NOT NULL DEFAULT '',
  `SUBJECT_NUM` varchar(10) NOT NULL DEFAULT '',
  `PRELIM` float(10,2) DEFAULT NULL,
  `MIDTERM` float(10,2) DEFAULT NULL,
  `FINALS` float(10,2) DEFAULT NULL,
  `FINAL_GRADE` float(3,2) DEFAULT NULL,
  PRIMARY KEY (`STUDENT_NUM`,`SUBJECT_NUM`))

CREATE TABLE IF NOT EXISTS `students` (
  `STUDENT_NUM` varchar(10) NOT NULL,
  `LASTNAME` varchar(50) DEFAULT NULL,
  `FIRSTNAME` varchar(50) DEFAULT NULL,
  `MIDDLENAME` varchar(50) DEFAULT NULL,
  PRIMARY KEY (`STUDENT_NUM`)
)

CREATE TABLE IF NOT EXISTS `subjects` (
  `SUBJECT_NUM` varchar(10) NOT NULL,
  `SUBJECT_TITLE` varchar(100) DEFAULT NULL,
  `DEPARTMENT` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`SUBJECT_NUM`)
)

any help will be greatly appreciated

 

[wagex]

did you delete the password from the first code box? that could be a problem if not lol

 

[pcarini]

Why are you including configdb.php in one of the files, but not in the other?

Try putting the following lines at the top of both prelimgrading.php and updateprelimgrade.php, below the opening <?php tag:

ini_set('display_errors', 1);
error_reporting(E_ALL);

Do any warnings or errors display in the browser when you do this?

EDIT: Another thing to try is removing the onsubmit = 'return checkform(this)'; from your <form> tag. I'm thinking that because the Javascript function checkform doesn't exist your code, Javascript is crashing right there and perhaps not submitting the form.

 

[raident30]

did you delete the password from the first code box? that could be a problem if not lol

i didn't put a password on my databse

Why are you including configdb.php in one of the files, but not in the other?

Try putting the following lines at the top of both prelimgrading.php and updateprelimgrade.php, below the opening <?php tag:

ini_set('display_errors', 1);
error_reporting(E_ALL);

Do any warnings or errors display in the browser when you do this?

EDIT: Another thing to try is removing the onsubmit = 'return checkform(this)'; from your <form> tag. I'm thinking that because the Javascript function checkform doesn't exist your code, Javascript is crashing right there and perhaps not submitting the form.

the warnings don't display, just my own warning "failed to update the grading".

i removed the javascript tag, its still not updating...

 

[rebelwarlock]

Replace every single one of your die() functions with die(mysql_error()). That should give you a better idea of where it's failing.

 

[raident30]

the error is here:

$query="update enrollment SET prelim=$prelim where student_num=$student_num";

its failing to update the database..

 

[pcarini]

The value for student_num needs to be wrapped in single quotes because it's a varchar.
Try:

$query="update enrollment SET prelim=$prelim where student_num='$student_num'";

 

[raident30]

The value for student_num needs to be wrapped in single quotes because it's a varchar.
Try:

$query="update enrollment SET prelim=$prelim where student_num='$student_num'";

oh my god it did work!!!! thank you very much!!!!!

 

[dropadrop]

BTW I would advise using more logging with different levels (debug, info, warn etc), can help solve problems like this faster... For example when debugging try to log the actual failed sql and then try entering it manually if you can't spot a problem with it.