SQL statement in php will not insert onto database

Question

I have this SQL statement that i'm trying to get to save new students to a table of students, however it simply isn't doing it, I don't get any error messages when I run error reporting and I ran the Query in sqlbuddy with values swapped in and it worked fine. Any ideas on what im doing wrong will be appreciated.

Heres the code:

<?php
session_start();
ini_set('display_startup_errors',1);
ini_set('display_errors',1);
error_reporting(-1);
$default = 'default';
$ClassID = $_GET['ID'];
$Surname = $_POST['Surname'];
$Firstname = $_POST['Firstname'];
$Firstletter = $Firstname[0];
$Username = $Firstletter + $Surname;
$sql_link = mysqli_connect('localhost', 'root', 'password', 'GameData');
$counter = mysqli_query($sql_link,"SELECT * FROM IDCounter");
$counter = mysqli_fetch_array($counter);
mysqli_query($sql_link,"INSERT INTO tblStudents(StudentID, StudentFirstName,    StudentSurname, ClassID, UserName, Password, CharacterSelect)
VALUES ('$counter[Counter]', '$_POST[Firstname]', '$_POST[Surname]', '$ClassID',  '$Username', '$default', 1)");
mysqli_close($sql_link);
header ("Location: TeacherSide.php");
?>

The POST values come from the form that directs to this page

Remove the header() call and add echo mysqli_error($sql_link); — Do you get any errors? — Amal, Commented Apr 28, 2014 at 19:42
Upon using the error check I got this error message: Warning: mysqli_error(): Couldn't fetch mysqli in /var/www/game/classmod.php on line 18 — Benjamin Sealy-Minto, Commented Apr 28, 2014 at 19:45
allow me to correct myself, I put it after the sql_close, apologise for any confusion, after putting it somewhere more suitable no issues were displayed — Benjamin Sealy-Minto, Commented Apr 28, 2014 at 20:09
When using mysqli you should be using parameterized queries and bind_param to add user data to your query. DO NOT use string interpolation to accomplish this because you will create severe SQL injection bugs. $_POST data never goes directly in the query. — tadman, Commented Apr 28, 2014 at 20:13
For reasons I dont understand the page temporarily worked, then I changed $_POST[Firstname] and $_POST[Surname] to there respective variable stated earlier in the code and now the statement fails to work again. — Benjamin Sealy-Minto, Commented Apr 28, 2014 at 20:21

Benjamin Sealy-Minto · Accepted Answer · 2014-04-28 20:24:06Z

0

I just worked out the issue I was having and I regret to inform you it was a rather stupid one, I was not updating my counter, So every time I tried to add a new student it would try with the same StudentID, and thus would fail, an easy fix

answered Apr 28, 2014 at 20:24

Benjamin Sealy-Minto

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