What is the mistake in this PHP code? - php

Question

I have a user registration system but its empty. This is the script I use in forum.modxpertz.tk. It worked at first but it shows nothing now. Here is the 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();
?>
I know only little about PHP and jQuery.

Answer 1

Please try below 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);
?>