Hi, I am having an error when tying to add data into my database. I have put up my code below as well as a link to a screenshot of my database. Any input is appreciated, thanks!

<?php

$studentfname = $_POST['studentfname'];
$studentlname = $_POST['studentlname'];
$course = $_POST['course'];
$gradecode = $_POST['gradecode'];
$gradep = $_POST['gradep'];


//database connection
$connection = new mysqli('localhost','root','root','grade');
if($connection->connect_error){
    echo "$connection->connect_error";
    die('Connection Failed :' .$connection->connect_error);}
    else{
        $stmt = $connection->prepare("insert into grade(studentfname, studentlname, course, gradecode, gradep) 
        values(?, ?, ?, ?, ?,)");
        $stmt->bind_param("ssssf", $studentfname,$studentlname,$course,$gradecode,$gradep);
        $stmt->execute();
        echo "Successful";
        $stmt->close();
        $connection->close();
    }

?>

https://imgur.com/a/n9EeJfB

    The error you are getting is a follow-on error because a previous statement failed, the ->prepare() call, but you don't have any error handling to tell you if and why a statement has failed, nor to stop code execution upon an error that will prevent the remainder of the code from running correctly.

    You ALWAYS need error handling for statements that can fail. For database statements that can fail - connection, query, prepare, and execute, the simplest way of adding error handling, without adding code at each statement, is to use exceptions for errors and in most cases let php catch and handle the exception, where php will use its error related settings to control what happens with the actual error information (database statement errors will 'automatically' get displayed/logged the same as php errors.) The exception to this rule is when inserting/updating duplicate user data, which is actually something your code needs to address. In this case, your code should catch the exception, detect if a duplicate index error (number) occurred, and setup a message telling the user what was wrong with the data that they submitted. For all other error numbers, just re-throw the exception and let php handle it.

    To enable exceptions for errors for the mysqli extension add the following line of code before the point where you make the connection (you can then remove any existing error handling logic since it won't get executed upon an error, simplifying the code) -

    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

    At a guess, it's the trailing comma in:

    values(?, ?, ?, ?, ?,)");
                    ^
                    |

    pbismad Sorry if i miss understood you but are you saying that I need to add mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); below my //database connection line?

      pbismad This has worked, I then had a couple other errors which where quite simple to fix, thanks

          EDIT: Was typing this up before I saw your post that you fixed it. 🙂

          You could also just directly test the prepare and see what it complains about:

          $stmt = $connection->prepare("insert into grade(studentfname, studentlname, course, gradecode, gradep) 
        values(?, ?, ?, ?, ?,)");
 if($stmt == false) {
     // just for debug for now, in production you'd just want to log this
     // and output something more user-friendly to the user 
     die("Prepare failed: ".$connection->error);
 }
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