PHP query not working to check database for username - php

Question

So I made a function that checks if the data that the user enters is already in my database of users. The function works in a testing environment (but runs on open of the page). For some reason it assigns the user input to the $username function regardless of the outcome of the query. I think I may have to AJAX it but I don't know how.
<?php
function checkIfEntered($data, $conn)
{
$query = "SELECT * FROM users
WHERE username= '" . $data . "'";
if ($result = mysqli_query($conn, $query)) {
/* determine number of rows result set */
$row_cnt = mysqli_num_rows($result);
/* close result set */
mysqli_free_result($result);
}
if (($row_cnt) >= 1)
{
return TRUE;
}
if (($row_cnt) == 0)
{
return FALSE;
}
}
error_reporting(E_ALL);
ini_set('display_errors',1);
// define variables and set to empty values
$servername = "localhost"; $username = "root";
$password = "root";
$dbname = "MyDatabase";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$email = '';
$emailErr = '';
$usernameErr = '';
$UserPasswordErr = '';
$username = '';
$UserPassword = '';
$confirm = '';
$confirmErr = '';
function test_input($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["username"]))
{
$usernameErr = "username is required";
}
else
{
if (checkIfEntered($username, $conn))
{
$usernameErr = "There is already a user with the username: " . $username;
}
else
{
$username = test_input($_POST["username"]);
}
}

Answer 1

You are not even calling the "checkIfEntered" function. Instead, you are calling "isEnteredUsername" function, which I don't see it here. So the code is always going in else and assigning the POST value to $username.
I am not sure why you think AJAXing it will solve the problem, but you can use AJAX call on document ready or some event like this $.ajax({ ... }) as explained here.
http://api.jquery.com/jquery.ajax/

Answer 2

you are forgot return num of record this function "checkIfEntered"
<?php
function checkIfEntered($data, $conn)
{
$query = "SELECT * FROM users
WHERE username= '" . $data . "'";
if ($result = mysqli_query($conn, $query)) {
/* determine number of rows result set */
$row_cnt = mysqli_num_rows($result);
/* close result set */
mysqli_free_result($result);
return $row_cnt;
}
if (($row_cnt) >= 1)
{
return TRUE;
}
if (($row_cnt) == 0)
{
return FALSE;
}
}
error_reporting(E_ALL);
ini_set('display_errors',1);
// define variables and set to empty values
$servername = "localhost"; $username = "root";
$password = "root";
$dbname = "MyDatabase";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$email = '';
$emailErr = '';
$usernameErr = '';
$UserPasswordErr = '';
$username = '';
$UserPassword = '';
$confirm = '';
$confirmErr = '';
function test_input($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["username"]))
{
$usernameErr = "username is required";
}
else
{
if (checkIfEntered(test_input($_POST["username"]), $conn))
{
$usernameErr = "There is already a user with the username: " . $username;
}
else
{
$username = test_input($_POST["username"]);
}
}
?>

Answer 3

The Problem with this code is the function checkIfEntered() is coming up false because when you pass the $username parameter it has not been assigned. so it is coming up as false.
Code:
if (checkIfEntered(test_input($_POST["username"]), $conn))
{
$usernameErr = "There is already a user with the username: " . $username;
}
else
{
$username = test_input($_POST["username"]);
}
Correction
$username = test_input($_POST["username"]);
if (checkIfEntered(test_input($_POST["username"]), $conn))
{
$usernameErr = "There is already a user with the username: " . $username;
$username = '';
//set username back to nothing to avoid more errors.
}
else
{
$username = test_input($_POST["username"]);
}