kyle_maddisson

First of all, thank you for your reply. Does that mean I need to make another variable and assign it the table name ($databasetable = database.table)? I assume you cannot connect to the table because I tried hard coding it there and would get access denied error. Once I added '.Students' to the assigned database name, it worked.

Can someone tell me why in the following code, in the 'try' section, this code works: (first line after 'try') $conn = new PDO("mysql:host=$servername;dbname=$dbasename", $username, $password); even though I have not declared a variable named $dbname, but if I change it to "$databasename", which I have declared, it doesn't work. I don't understand that. <code> <?php $databasename = "removed"; $servername = "removed"; $username = "removed"; $password = "removed"; $id=$_POST[id]; $courseID=$_POST[courseID]; $coursename=$_POST[coursename]; $firstname=$_POST[firstname]; $lastname=$_POST[lastname]; $middlename=$_POST[middlename]; $grade=$_POST[grade]; $year=$_POST[year]; $sex=$_POST[sex]; echo "<br/>Connecting to Database.....<br/>"; try { $conn = new PDO("mysql:host=$servername;dbname=$dbasename", $username, $password); // set the PDO error mode to exception $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $sql = "INSERT INTO $databasename (studentID, lastName, firstName, middleName, courseName, courseNumber, year, grade, gender) VALUES ('$id','$lastname','$firstname','$middlename','$coursename','$courseID','$year','$grade', '$sex')"; $conn->exec($sql); echo "New record created successfully"; } catch(PDOException $e) { echo $sql . "<br>" . $e->getMessage(); } $conn = null; ?> </code>