You have an error in your SQL syntax, why? - php

Question

My PHP code.
<?php
$servername = "localhost";
$username = "xxxxxxxxx";
$password = "xxxxxxxxx";
$database = "xxxxxxxxx";
$postdata = file_get_contents("php://input");
$data = json_decode($postdata, true);
$table = $data['table'];
$name = $data['name'];
$email = $data['email'];
$points = $data['points'];
$percentage = $data['percentage'];
$userAnswers = $data['userAnswers'];
function connect_DB($servername, $username, $password, $database) {
$db_conn = new mysqli($servername, $username, $password, $database);
$char = $db_conn->query("SET NAMES 'utf8'");
if ($db_conn->connect_error) {
header($_SERVER['SERVER_PROTOCOL'] . ' 500 Internal Server Error', true, 500);
die("Connection failed: " . $db_conn->connect_error);
}
else {
echo "Connected successfully \r\n";
}
return $db_conn;
}
function create_DB_table($db, $table) {
$sql = "CREATE TABLE IF NOT EXISTS $table (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
reg_date TIMESTAMP
)";
if ($db->query($sql) == TRUE) {
echo "Table successfully created \r\n";
}
else {
header($_SERVER['SERVER_PROTOCOL'] . ' 500 Internal Server Error', true, 500);
die("Table creation failed: " . $db->error);
}
}
function addColumnVarchar($conn_obj, $table, $column) {
$sql = $conn_obj->prepare("SHOW COLUMNS FROM $table LIKE '%$column%'"); // add wildcard
$sql->execute();
if($sql->num_rows <= 0) {
$sql->store_result();
$sql1 = $conn_obj->prepare("ALTER TABLE $table ADD COLUMN $column VARCHAR(255)");
$sql1->execute();
}
}
function addColumnFloat($conn_obj, $table, $column) {
$sql = $conn_obj->prepare("SHOW COLUMNS FROM $table LIKE '%$column%'"); // add wildcard
$sql->execute();
if($sql->num_rows <= 0) {
$sql->store_result();
$sql1 = $conn_obj->prepare("ALTER TABLE $table ADD COLUMN $column FLOAT");
$sql1->execute();
}
}
function insert_DB($db, $table, $column, $value) {
$sql = "INSERT INTO $table ($column)
VALUES ($value)";
if( $db->query($sql) == TRUE) {
echo "Records inserted successfully!";
}
else {
header($_SERVER['SERVER_PROTOCOL'] . ' 500 Internal Server Error', true, 500);
die("Records insertion failed: " . $db->error);
}
}
//connect to the database and create table
$conn_obj = connect_DB($servername, $username, $password, $database);
create_DB_table($conn_obj, $table);
$columnArr = array();
$valueArr = array();
if (!is_null($name)){
addColumnVarchar($conn_obj, $table, 'name');
array_push($columnArr, "name");
array_push($valueArr, $name);
}
if (!is_null($email)){
addColumnVarchar($conn_obj, $table, 'email');
array_push($columnArr, "email");
array_push($valueArr, $email);
}
if (!is_null($points)){
addColumnFloat($conn_obj, $table, 'points');
array_push($columnArr, "points");
array_push($valueArr, $points);
}
if (!is_null($percentage)){
addColumnFloat($conn_obj, $table, 'percentage');
array_push($columnArr, "percentage");
array_push($valueArr, $percentage);
}
if (!is_null($userAnswers)){
foreach ($userAnswers as $ua) {
addColumnVarchar($conn_obj, $table, $ua['qID']);
array_push($columnArr, $ua['qID']);
array_push($valueArr, wordwrap($ua['answer'], 60, "\n", false));
}
}
$column = implode(",", $columnArr);
$value = "'".implode("','", $valueArr)."'";
insert_DB($conn_obj, $table, $column, $value);
$conn_obj->close();
The output is:
Connected successfully Table creation failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '( id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, reg_date TIMEST' at line 1
Is there a syntax error?