Was ist der Fehler in diesem PHP-Code?

Question

Ich habe ein Benutzerregistrierungssystem, aber es ist leer. Dies ist das Skript, das ich in forum.modxpertz.tk verwende. Es hat zuerst funktioniert, aber es zeigt jetzt nichts. Hier ist der Code.

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = ""; 

// Create connection 
$conn = mysqli_connect($servername, $username, $password); 
mysqli_select_db($conn,'login'); 

// Check connection 
if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 

$sql = "SELECT userid FROM login"; 
$result = mysqli_query($conn, $sql); 

if (mysqli_num_rows($result) > 0) { 
// output data of each row 
while($row = mysqli_fetch_assoc($result)) { 
    $reguserid=$row["userid"]; 
} 

$userid = mysqli_real_escape_string($conn, $_POST['userid']); 
$pswrd = mysqli_real_escape_string($conn, $_POST['pswrd']); 
$fname = mysqli_real_escape_string($conn, $_POST['fname']); 
$lname = mysqli_real_escape_string($conn, $_POST['lname']); 
$gender = mysqli_real_escape_string($conn, $_POST['gender']); 

$token = rand('122332344','922332344'); 
$url = array('forum.modzexpertz.tk/verify.php#',$token); 

$post= join($url); 

if($userid!=$reguserid){ 
    $sql = "INSERT INTO login(fname, lname, userid, pswrd, gender)VALUES('$fname', '$lname', '$userid', '$pswrd', '$gender')"; 

if ($conn->query($sql) === TRUE) { 
    echo "New record created successfully"; 
}else { 
echo "Failed to Register."; 
}} else { 
echo "A user with the email youve provided has already been registered."; 
}} 
$conn->close(); 
?> 

Ich weiß nur wenig über PHP und jQuery.

Answer 1

Bitte versuchen Sie es unten Code:

<?php 
$servername = "localhost"; 
$username = "root"; 
$pswrd = ""; 
$db = "login"; 

$conn = mysqli_connect($servername,$username,$pswrd, $db); 

// Check connection 
if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$table = 'login'; 
if(@mysqli_num_rows(mysqli_query($conn, "SELECT NULL FROM `$table` WHERE userid='".$_POST['userid']."'")) > 0){ 
      $error = "1"; 
      echo "user with same userid is already exist"; 
} 

if(isset($_POST['fname']) && isset($_POST['lname']) && isset($_POST['gender']) && isset($_POST['userid']) && isset($_POST['pswrd']) && $_POST['fname']!="" && $_POST['lname']!="" && $_POST['gender']!="" && $_POST['userid']!="" && $_POST['pswrd']!="") 
{ 
    if($error==''){ 
     $ins['fname'] = mysqli_real_escape_string($conn, $_POST['fname']); 
     $ins['lname'] = mysqli_real_escape_string($conn, $_POST['lname']); 
     $ins['gender'] = mysqli_real_escape_string($conn, $_POST['gender']); 
     $ins['userid'] = mysqli_real_escape_string($conn, $_POST['userid']); 
     $ins['pswrd'] = mysqli_real_escape_string($conn, $_POST['pswrd']); 
     $insertsql = "INSERT INTO `$table` (fname, lname, gender, userid, pswrd) VALUES ('".$ins['fname']."','".$ins['lname']."','".$ins['gender']."','".$ins['userid']."','".$ins['pswrd']."')"; 
     @mysqli_query($conn, $insertsql); 
     //echo $insertsql; exit; 
     echo "Success"; 
    } 
}else{ 
    echo "Please enter required parameters"; 
} 

mysqli_close($conn); 
?>