Question

I have a block of code below which inserts data into database using mysqli and php. The problem is that I am getting a fatal error stating: Fatal error: Cannot pass parameter 2 by reference in … on line 116

Why is this error appearing and how can I fix the error?

Below is the code:

if ($numrows == 0){

   $teacherpassword = md5(md5("j3Jf92".$teacherpassword."D203djS"));  
   $code = md5(rand());

    $insertsql = "
    INSERT INTO Teacher
        (TeacherId, TeacherForename, TeacherSurname, TeacherEmail, TeacherAlias, TeacherUsername, TeacherPassword, Active, Code)
      VALUES
        (?, ?, ?, ?, ?, ?, ?, ?, ?)
    ";
    if (!$insert = $mysqli->prepare($insertsql)) {
      // Handle errors with prepare operation here
    }                                           

    $insert->bind_param("sssssssss", '', $getfirstname, $getsurname,
                 $getemail, $getid, $getuser,
                 $teacherpassword, '0', $code);

    $insert->execute();

    if ($insert->errno) {
      // Handle query error here
    }

    $insert->close();

Answer 1

mysqli_stmt::bind_param() takes one string detailing the types of the following arguments, and then a set of references to variables that contain the data.

Only variables may be passed by reference, so you are not allowed to pass a string ('' or '0') to the function. You must put that string in a variable, and then pass that variable.

If you’re passing constant values to an INSERT, why not make them the DEFAULT values of those fields and then remove them from the query?