Question

I am building a simple website that helps students and instructors in universities.

I am facing a problem about the following query:
An instructor retrieves the students’ IDs and names who are enrolled in the course he/she teaches.

I have the following tables, followed by their fields:
Enrollment (CourseCode – StudentID – Grade)
Studnet (ID – Name)

As you can see the only connector between the two tables is the student ID.

The code that I wrote is

<?
session_start();
$COCODE = $_SESSION['GlobalCode'];

$result11 = mysql_query("SELECT * FROM Enrollment WHERE CourseCode = '$COCODE' ") ;

$row11 = mysql_fetch_array($result11);
$StID = $row11['StudentID'];
$result22 = mysql_query("SELECT * FROM Student where StudentID= '$StID' ") ;
echo "<table border cellpadding=3>";

while($row123 = mysql_fetch_array($result22)) 
 { 
 echo "<tr>";
 echo "<td>".$row123['ID']."</td> ";
 echo "<td>".$row123['Name']."</td> ";
 echo "</tr>";
 } 

echo "</table>"; 
?>

What I am trying to do is to retrieve the course code from the Enrollment table and then retrieving the students names through the ID.

The problem is that I got the following message:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

I hope you can help me solving the problem.

Thanks

Answer 1

Have you tried to narrow down which of the queries is causing the problem? That would be your first step. Some other pointers:

1. you need to check that each SQL query is successful and returning a valid value before using it in the next query… otherwise that query will cause an error.

2. Try using: mysql_query ($your_query) or die ('Error: '.mysql_error ()); for each query for a more detailed error message. (for debugging only, not for production)