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  1. #1

    Thread Starter
    Addicted Member
    Join Date
    Nov 2004
    Posts
    144

    Varibles

    I'm trying to access a variable from another form but cant seem to get it to work can any one see my problem?


    Main Menu
    Code:
    <?
    // script to display all the Degrees in the Degree table
    
    // connection information
    $hostName = "localhost";
    $userName = "root";
    $password = "12345";
    $dbName = "SOC_DBtest";
    
    // make connection to database
    mysql_connect($hostName, $userName, $password) or die("Unable to connect to host $hostName");
    
    mysql_select_db($dbName) or die("Unable to select database $dbName"); 
    
    // Select all the fields in all the records of the Employees table
    $query = "SELECT *
              FROM tbl_degree";
    
    $result = mysql_query($query);
    
    // Determine the number of employees
    $number = mysql_numrows($result);
    
    // Create drop-down menu of degree names 
    
    print "View students enroled on a specified degree scheme:<p>
        <form action=\"StudentDegree.php\" method=\"post\">
        <select name=\"DegreeId\">
        <option value=\"\">Select a degree scheme</option>";
    
    for ($i=0; $i<$number; $i++) {
         $DegreeId  = mysql_result($result,$i,"DegreeId");
         $Degree_Name = mysql_result($result,$i,"Degree_Name");
         print "<option value=\"$DegreeId\">$Degree_Name</option>";
    }
    
    print "</select><input type=\"submit\" value=\"submit\"
        name=\"submit\"></form>";
    
    // Close the database connection
    mysql_close();
    
    ?>
    Result Page
    Code:
    <?
    // script to display who is on a specific degree
    
    // connection information
    $hostName = "localhost";
    $userName = "root";
    $password = "12345";
    $dbName = "SOC_DBtest";
    
    
    // make connection to database
    mysql_connect($hostName, $userName, $password) or die("Unable to connect to host $hostName");
    
    mysql_select_db($dbName) or die( "Unable to select database $dbName"); 
    
    
    // Select the fields from the appropriate tables
    $query =
    
    
    "SELECT tbl_student.Student_Id, tbl_student.Student_FName, tbl_student.Student_SName, tbl_degree.Degree_Name
    FROM tbl_degree INNER JOIN tbl_student ON tbl_degree.Degree_Id = tbl_student.Degree_Id
    WHERE ((tbl_student.Degree_Id = $DegreeId))";
    
    $result = mysql_query($query);
    
    // Determine the number of records returned
    $number = mysql_numrows($result);
    
    // Print the relevant information
    
    print "There are $number student(s) studying this Degree:<p>";
    
    for($i=0; $i<$number; $i++) {
         $studentId = mysql_result($result, $i, "student_Id");
         $stud_FName = mysql_result($result, $i, "Student_FName");
         $stud_SName = mysql_result($result,$i,"Student_SName");
         $stud_Degree = mysql_result($result,$i,"Degree_Name");
         print "$studentId - $stud_FName $stud_SName - $stud_Degree<br>";
    }
    
    // Close the database connection
    mysql_close();
    ?>
    it prings up an error with my SQL query, i think it's the WHERE clause, where the $DegreeID isn't being accessed from the main menu
    Last edited by JamesBowtell; Mar 3rd, 2006 at 12:03 PM.

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