PHP - Sending Textbox Data To Database Causing Two Elements To Disappear. Help?

Hi,

For about a month, I have been trying to figure out why my code will not return anything after posting a wwwForm (I have also tried the newer equivalent of this function but I had no luck with that either.) The nameField and passwordField are taken from text boxes within the game and the code used in my login script is copied and pasted from a Register script but I have changed the file location to the login.php file. The register script works fine and I can add new users to my database but the login script only outputs "Form Sent." and not the "present" that should return when the form is returned and it never gets any further than that point meaning that it lets the user through with no consequence if they use an invalid name because the script never returns an answer. What should I do to fix this?

Thanks,

Unity Code:

using System.Collections;
using UnityEngine;
using UnityEngine.UI;
using UnityEngine.Networking;

public class Login : MonoBehaviour
{
    public InputField nameField;
    public InputField passwordField;

    public Button acceptSubmissionButton;

    public void CallLogInCoroutine()
    {
        StartCoroutine(LogIn());
    }

    IEnumerator LogIn()
    {
        WWWForm form = new WWWForm();
        form.AddField("username", nameField.text);
        form.AddField("password", passwordField.text);
        WWW www = new WWW("http://localhost/sqlconnect/login.php", form);
        Debug.Log("Form Sent.");
        yield return www;
        Debug.Log("Present");        if (www.text[0] == '0')
        {
            Debug.Log("Present2");
            DatabaseManager.username = nameField.text;
            DatabaseManager.score = int.Parse(www.text.Split('\t')[1]);
            Debug.Log("Log In Success.");
        }
        else
        {
            Debug.Log("User Login Failed. Error #" + www.text);
        }
        
    }

    public void Validation()
    {
        acceptSubmissionButton.interactable = nameField.text.Length >= 7 && passwordField.text.Length >= 8;
    }
}

login.php:

<?php
    echo "Test String2";
    $con = mysqli_connect('localhost', 'root', 'root', 'computer science coursework');

    // check for successful connection.
    if (mysqli_connect_errno())
        {
            echo "1: Connection failed"; // Error code #1 - connection failed.
            exit();
        }

    $username = mysqli_escape_string($con, $_POST["username"]);
    $usernameClean = filter_var($username, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW | FILTER_FLAG_STRIP_HIGH);
    $password = $_POST["password"];

    if($username != $usernameClean)
    {
        echo "7: Illegal Username, Potential SQL Injection Query. Access Denied.";
        exit();
    }

    // check for if the name already exists.

    $namecheckquery = "SELECT username, salt, hash, score FROM players WHERE username='" . $usernameClean . "';";

    $namecheck = mysqli_query($con, $namecheckquery) or die("2: Name check query failed"); // Error code # 2 - name check query failed.

    if (mysqli_num_rows($namecheck) != 1)
    {
    	echo "5: No User With Your Log In Details Were Found Or More Than One User With Your Log In Details Were Found"; // Error code #5 - other than 1 user found with login details
    	exit();
    }

    // get login info from query
    $existinginfo = mysqli_fetch_assoc($namecheck);
    $salt = $existinginfo["salt"];
    $hash = $existinginfo["hash"];

    $loginhash = crypt($password, $salt);
    if ($hash != $loginhash)
    {
    	echo "6: Incorrect Password"; // error code #6 - password does not hash to match table
    	exit;
    }
    echo "Test String2";
    echo"0\t" . $existinginfo["score"];

?>

I am building a system where people are able to choose a food menu, they can modify each dish if needed and then say how many of each item they need for the order.

So when they pick the breakfast menu, it selects all starters, mains and desserts.  Should something need to be edited (sausages changed to vegitarian sausages) the user should be able to overtype what has been populated from the select statement)

When the form is submitted, the order needs to be linked to the current job which is in a field ssm_job.job_id but the changes to the menu items need to be stored somewhere for the current order only.  This obviously needs to be accessile later so further changes can be made if needed.  I am not sure of the best way to go about this as i am trying not to save every item in another database (just trying to be tidy)

Any advise on this would be greatly appreciated.

I have had help on this project in relation to an sql and the background can be found below, the help has been amazing but as this is a very different question i thought i would start a new thread.

Kind Regards

Hi all,

I have made some code based on following online youtube video for making a HTML form send to mySQL database via PHP: https://youtu.be/qm4Eih_2p-M

The form I am making is hopefully going to help in the Hospital at which I work to speed patient referrals, so I am very keen on making it work.

I have made the HTML form and the database, which I do not think are the problem.

When I click 'Submit' on my HTML form, then my php file comes up with all the code written, rather than the message: "New referral sent succesfully".

Additionally, when I check the database via 'MyPHPAdmin' there is no data added to the database I have made.

I have been using Sublime and have spotted some mistakes here and there, but is there something glaringly obvious that I am doing wrong in my code that is stopping my form being sent to the database?

Thank you,
Saman

<?php

$patientname = | $_POST['patientname'];
$hospitalnumber = | $_POST['hospitalnumber'];
$DoB = | $_POST['DoB'];
$specialty = | $_POST['specialty'];
$otherspecialty = | $_POST['otherspecialty'];
$admissiondate = | $_POST['admissiondate'];
$referring = | $_POST['referring'];
$bleep = | $_POST['bleep'];
$Admission = | $_POST['Admission'];
$Illness = | $_POST['Illness'];
$Awareness = | $_POST['Awareness'];
$question = | $_POST['question'];
$history = | $_POST['history'];
$medications = | $_POST['medications'];
$examination = | $_POST['examination'];
$results = | $_POST['results'];
$diagnosis = | $_POST['diagnosis'];
$investigations = | $_POST['investigations'];

	$host = "localhost"
	$dbUsername = "root";
	$dbPassword = "xyzabc";
	$dbname = "GreenCard";

	$conn = new mysqli($host, $dbUsername, $dbPassword, $dbname);

	if (mysqli_connect_error()) {
		die('Connect Error('. mysqli_connect_errno().')'. mysqli_connect_error());
	
	} else {
	$INSERT = "INSERT Into Greencard (patientname, hospitalnumber, DoB, specialty, otherspecialty, admissiondate, referring, bleep, Admission, Illness, Awareness, question, history, medications, examination, results, diagnosis, investigations) values (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";

	//Prepare statement
	$stmt = $conn->prepare($INSERT);
	$stmt->bind_param("sisissssssssssssss", $patientname, $hospitalnumber, $DoB, $specialty, $otherspecialty, $admissiondate, $referring, $bleep, $Admission, $Illness, $Awareness, $question, $history, $medications, $examination, $results, $diagnosis, $investigations);
	$stmt->execute();
	echo "New referral sent successfully";
	}
	$stmt->close();
	$conn->close();

} else {
	echo "Make sure fields are completed";
	die();
}

?>

                    echo"Auto/Prijs<br><br><select name='autos'>";
                echo"<br><br>";
                        
                $sql = "SELECT `garage`.`id`, `car_id`, `schade`, `naam`, `prijs` FROM `garage` LEFT JOIN `cars` ON (`garage`.`car_id` = `cars`.`id`) WHERE `user_id`=".ID." ORDER BY `id` ASC LIMIT ".($page * 10).", 10";
    $sql = mysql_query($sql) or die(mysql_error());
        $i = 1;
    while($res = mysql_fetch_assoc($sql)){
        echo"
        
    
    <option value='".$res['car_id']."'>".$res['naam']."</option><br>
        ";

    if(isset($_POST['start'])){
    $prijs = $_POST['prijs'];
    $carr = $_POST['autos'];
    $sql = mysql_query("SELECT `id` FROM `automarkt` WHERE `seller_id`=".ID." LIMIT 1") or die(mysql_error());
    mysql_query("INSERT INTO `automarkt`(`seller_id`, `prijs`, `car_id`) VALUES (".ID.", ".$prijs.", ".$carr.")") or die(mysql_error());