Mysqli prepared statements error? [duplicate] - php

This question already has answers here:
mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement
(2 answers)
Closed 1 year ago.
I've ran into this error with prepared statements, I've just started with prepared statements so go easy on me please, Heres the error:
Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in C:\wamp\www\darkhorizons\login.php on line 31
Heres my code:
if (isset($_POST['submit'])){
$username = $_POST['username'];
$password = $_POST['password'];
if(isset($username) && isset($password)) {
$mysqli = new mysqli("localhost","root","","phplogin") or die("Couldnt connect!");
if(mysqli_connect_errno()){
echo "Connection failed: ". mysqli_connect_errno();
exit();
}
if($stmt = $mysqli -> prepare("SELECT * FROM users WHERE username =? AND password =? LIMIT 1")){
$stmt -> bind_param("ss", $username, $password);
$stmt -> execute();
$stmt -> bind_result($result);
$stmt -> fetch();
$numrows = mysqli_num_rows($result);
} else {
die("Please enter a username and password!");
}
if($numrows == 1){
$_SESSION['username'] = $_POST['username'];
$_SESSION['loggedin'] = true ;
$query = "SELECT adminflag FROM users WHERE username = '{$_SESSION['username']}' LIMIT 1;";
$result2 = mysqli_query($connect, $query);
$numrows2 = mysqli_num_rows($result2);
if ($numrows2 == 1) {
$_SESSION['isadmin'] = true;
}
header("Location: {$pageLoc}");
exit(); //It's good to use exit or die (same thing) AFTER using header to redirect
} else {
}
}
}
As a side note, Please ignore any mistakes in the code below the prepared statement, im redoing my login script that ive been using to learn.

Going through your code you didn't really need to query you DB twice, you should read the adminflag in that same select.
SELECT * is never a good idea always select specific fields.
And I also noticed you are using two differnt style, I suggest you to stick to the Object oriented approach.
<?php
if (isset($_POST['submit'], $_POST['username'] , $_POST['password'])){
$username = $_POST['username'];
$password = $_POST['password'];
$mysqli = new mysqli("localhost","root","","phplogin");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "SELECT adminflag FROM users WHERE username = ? AND password = ? LIMIT 1";
if ($stmt = $mysqli->prepare($query)) {
$stmt -> bind_param("ss", $username, $password);
$stmt->execute();
$stmt->store_result();
$numrows = $stmt->num_rows;
printf("Number of rows: %d.\n", $numrows );
if($numrows == 1){
$stmt->bind_result($admin_flag);
$stmt->fetch();
session_start();
if ($admin_flag== 1) {
$_SESSION['isadmin'] = true;
}
$_SESSION['username'] = $username;
$_SESSION['loggedin'] = true ;
header("Location: {$pageLoc}");
}else{
echo 'user not found';
}
}
$stmt->close();
$mysqli->close();
}else{
echo 'required field missing';
}
?>

Related

password_verify having a few issues hopefully someone can help me

So im making a login for a game system that uses a JSON string to read back the files, however when I Hash the password and try login it doesn't work however when the password is unhased it works fine, I think the problem lies with the password verify part.
Im just unsure where to place it if someone could guide me in the right direction that will be amazing...
So the issue is when I send the $password example test5 it only reads as test 5 and not this $2y$10$viov5WbMukXsCAfIJUTUZetGrhKE9rXW.mAH5F7m1iYGfxyQzQwD.
This was the original Code
<?php
include("dbconnect.php");
/////// First Function To Get User Data For Login
if($_GET["stuff"]=="login"){
$mysqli = new mysqli($DB_host, $DB_user, $DB_pass, $DB_name);
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/////// Need to Grab Username And Password
$username = $_GET["user"];
$password = $_GET["password"];
$query = "SELECT username, password, regkey, banned FROM users WHERE username='$username' and password='$password'";
if ($stmt = $mysqli->prepare($query)) {
$stmt->execute();
$stmt->bind_result($username, $password, $regkey, $banned);
while ($stmt->fetch()) {
echo "{";
echo '"result": "success",';
echo '"regkey": "' . $regkey . '",';
echo '"banned": "' . $banned . '"';
echo "}";
}
$stmt->close();
}
$mysqli->close();
}
Then I tried This
<?php
include("dbconnect.php");
/////// First Function To Get User Data For Login
if($_GET["stuff"]=="login"){
$mysqli = new mysqli($DB_host, $DB_user, $DB_pass, $DB_name);
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/////// Need to Grab Username And Password
$username = $_GET["user"];
$password = $_GET["password"];
$GetPassword = "SELECT password FROM users WHERE username='$username'";
$data = mysqli_query($GetPassword);
if(password_verify($password, $data['password'])) {
$query = "SELECT username, password, regkey, banned FROM users WHERE username='$username' and password='$password'";
if ($stmt = $mysqli->prepare($query)) {
$stmt->execute();
$stmt->bind_result($username, $password, $regkey, $banned);
while ($stmt->fetch()) {
echo "{";
echo '"result": "success",';
echo '"regkey": "' . $regkey . '",';
echo '"banned": "' . $banned . '"';
echo "}";
}
$stmt->close();
}
} else {
// password is in-correct
}
$mysqli->close();
}
$data = mysqli_query($GetPassword);
^^^---mysqli result handle/object
if(password_verify($password, $data['password'])) {
^^^^^^^^^^^^---no such element in mysqli
You never fetched your data results, so you're comparing the entered password against a non-existent element of the $data result object/handle.
You need
$row = mysqli_fetch_assoc($data);
password_verify(..., $row['password']);
Most likely you're running with error_reporting and display_errors disabled, meaning you'd never see the "undefined index" warnings that PHP would ave been throwing everytime you ran this code. Those settings should NEVER be off on a devel/debug system.
And note that you're vulnerable to sql injection attacks, so your "security" system is anything but -it's entirely useless and trivially bypassable.
There are several issues with your code, such as:
This statement $data = mysqli_query($GetPassword) is wrong. mysqli_query() expects first argument to be your connection handler, so it should be,
$data = mysqli_query($mysqli, $GetPassword);
Look at this statement, if(password_verify($password, $data['password'])) { ...
mysqli_query(), on success, returns a result set, so you can't get the password using $data['password']. First fetch the row from the result set and then get the password, like this,
$row = mysqli_fetch_assoc($data);
if(password_verify($password, $row['password'])) { ...
Look at the following query,
$query = "SELECT username, password, regkey, banned FROM users WHERE username='$username' and password='$password'";
This query won't return any rows because of this WHERE condition, ...password='$password'. So the solution is, instead using two separate queries, use only one query to validate the password and retrieve relevant data. The solution is given down below.
Your queries are susceptible to SQL injection. Always prepare, bind and execute your queries to prevent any kind of SQL injection.
If you want to build a json string then instead of doing echo "{"; echo '"result": "success",'; ..., create an array comprising of all the relevant data and then json_encode the array.
So the solution would be like this:
// your code
$username = $_GET["user"];
$password = $_GET["password"];
// create a prepared statement
$stmt = mysqli_prepare($mysqli, "SELECT username, password, regkey, banned FROM users WHERE username=?");
// bind parameters
mysqli_stmt_bind_param($stmt, 's', $username);
// execute query
mysqli_stmt_execute($stmt);
// store result
mysqli_stmt_store_result($stmt);
// check whether the SELECT query returned any row or not
if(mysqli_stmt_num_rows($stmt)){
// bind result variables
mysqli_stmt_bind_result($stmt, $username, $hashed_password, $regkey, $banned);
// fetch value
mysqli_stmt_fetch($stmt);
if(password_verify($password, $hashed_password)){
echo json_encode(array('result' => 'success', 'regkey' => $regkey, 'banned' => $banned));
}else{
// incorrect password
}
}else{
// no results found
}
mysqli_stmt_close($stmt);
// your code

Checking if Username and Password are correct

As my code is right now, I always get the echo "Username/Password incorrect."; whether or not the username/password match or not. My question is, What did I do wrong in the code below for the php to always echo "Username/Password incorrect"
<?php
require 'privstuff/dbinfo.php';
$password1 = $_POST["password1"];
$username = $_POST["username"];
$mysqli = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_DATABASE);
if(mysqli_connect_errno()) {
echo "Connection Failed. Please send an email to owner#othertxt.com regarding this problem.";
exit();
}
if ($stmt = $mysqli->prepare("SELECT username, password FROM accounts WHERE username=? and password=?")) {
$db_pw = password_hash($password1, PASSWORD_BCRYPT);
$stmt->bind_param("ss", $username, $db_pw);
$stmt->execute();
if ($stmt->affected_rows > 0) {
echo "Logged in.";
}else{
echo "Username/Password incorrect.";
}
$stmt->close();
}
$stmt->close();
$mysqli->close();
?>
Update I've changed if ($stmt->affected_rows > 0) to if ($stmt->num_rows). Still doesn't work though
UPDATE 2 I've realized the issue is me using password_hash($password1, PASSWORD_BCRYPT); I didn't realize that the hash gives different strings every time. I'm not understanding on how to use password_verify
The documentation of mysqli_stmt_affected_rows() says:
This function only works with queries which update a table. In order to get the number of rows from a SELECT query, use mysqli_stmt_num_rows() instead.
You also need to call mysqli_stmt_store_results() first, to buffer the results.
$stmt->store_results();
if ($stmt->num_rows > 0) {
...
}
I figured it out. I was not supposed to use password_hash again. I didn't realize that using password_hash gave different results. I then changed it to use password_verify.
<?php
require 'privstuff/dbinfo.php';
$username = $_POST["username"];
$password1 = $_POST["password1"];
$mysqli = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_DATABASE);
// Check connection
if(mysqli_connect_errno()) {
echo "Connection Failed: " . mysqli_connect_errno();
exit();
}
/* create a prepared statement */
if ($stmt = $mysqli->prepare("SELECT `password` FROM `accounts` WHERE username = ?")) {
/* Bind parameters: s - string, b - blob, i - int, etc */
$stmt -> bind_param("s", $username);
/* Execute it */
$stmt -> execute();
/* Bind results */
$stmt -> bind_result($result);
/* Fetch the value */
$stmt -> fetch();
/* Close statement */
$stmt -> close();
}
if(password_verify($password1, $result))
{
echo("Hello");
}else{
echo("No-Go");
}
$mysqli->close();
?>

PHP/MySQL: Check if username exists

I'm a beginner in php and I want to check if the username entered already exists.
Here is my code.
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
if (isset($_POST['submit'])) {
include "connect.php";
ValidateUser();
}
function ValidateUser()
{
if (!empty($_POST['username']) AND !empty($_POST['password'])) {
$queryrow=mysqli_query("SELECT * FROM websiteusers WHERE username = '$_POST['username']'");
if ($rows=mysqli_num_rows($queryrow)=0) {
RegisterUser();
}
}
function RegisterUser() {
echo "works up to here";
}
?>
It doesn't even give me an error despite turning error reporting on.
Have you even initialized a mysqli_connect?
$Connection = mysqli_connect("host","user","pass","database");
Then pass it to a function which uses mysqli_query() by:
function foo ($DB){
mysqli_query($DB,"QUERY HERE");
// Do other stuff
return /* Whatever you wish to return here*/
}
foo($Connection);
What you are trying to achieve can be done very easily with the following code. A bigger concern is security. It is good practice to both sanitize your input every time the user has a chance to input text.
Also, using prepared query's will put yet another layer of security.
Although this isn't using your provided code directly, I believe it is good to teach good habits.
If you have any questions feel free to ask.
$username = $_POST['username']; <-- sanitize this
$message = null;
$mysqli = new mysqli("localhost", "user", "password", "database");
$stmt = $mysqli->prepare("SELECT username FROM websiteusers WHERE username=?");
$stmt->bind_param('s', $username);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($usernamesql);
$stmt->fetch();
if ($stmt->num_rows() > 0) {
RegisterUser();
} else {
$message .= 'username already exists';
}
Later on when you require more items to be queried, or more results to be bound:
$stmt = $mysqli->prepare("SELECT username,password,other1,other2 FROM websiteusers WHERE username=?");
$stmt->bind_param('s', $username); <-- the "s" means the argument is a strings, if a argument is stored as an int use "i", but one character for each argument is required.
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($usernamesql);
$stmt->fetch();
Multiple Arguments:
$stmt = $mysqli->prepare("SELECT username,password,other1,other2 FROM websiteusers WHERE username=? AND authenticated=?");
$stmt->bind_param('si', $username,$isauthenticated); <-- second argument is a INT or BOOL
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($usernamesql,$passwordsql,$other1sql,$other2sql);
$stmt->fetch();
When your expecting multiple results, and lets say you want to dump them into arrays:
$userarray = array();
$stmt = $mysqli->prepare("SELECT username FROM websiteusers WHERE username=?");
$stmt->bind_param('s', $username);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($usernamesql);
while($stmt->fetch()){
array_push($userarray, $usernamesql);
}
$userarray is now an array of all the results fetched from the database.
Here is the right way to do this:
<?php
ini_set('display_errors',1);
error_reporting(E_ALL);
if (isset($_POST['submit'])) {
$mysqli = new mysqli("localhost", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
if(check_user($mysqli, $_POST['username']){
registerUser();
}else{
echo 'user exist, cannot register';
}
}
function check_user($conn, $username){
$query = "SELECT * FROM websiteusers WHERE username = ?";
if ($stmt = $conn->prepare($query)) {
$stmt->bind_param("s", $username);
$stmt->execute();
$stmt->close();
}
return $stmt->num_rows === 0;
}
function registerUser() {
echo "registering user ...";
}
Read up on prepared statement

Inserting data to mysql not working

<?
session_start();
if(($connection = mysql_connect("localhost", "root", "")) == false)
die ("Couldn't connect to database");
if(mysql_select_db("Social", $connection) == false)
die ("Couldn't select db");
if (isset($_POST['username']) && isset($_POST['pass']) && isset($_POST['login'])){
$sql = sprintf("SELECT * FROM users WHERE username LIKE '%s' AND password LIKE '%s'", $_POST['username'], $_POST['pass']);
$query = mysql_query($sql);
if (mysql_num_rows($query) == 0){
$error = "<br />Wrong Username or Password";}
else{
$_SESSION['user'] = $_POST['username'];
header("Location: home1.php");
}
}
if (isset($_POST['register'])){
$sql2 = sprintf("INSERT INTO Social.users (username, password) VALUES (%s, %s)", $_POST['newUser'], $_POST['newPass']);
$query2 = mysql_query($sql2);
if (!$query2){
print "Registration failed";
}
else{
print "Registration sucessfull";
}
}
?>
My program is not inserting any data into mySQL table. I know all the syntax is right, everything should work out fine. I double checked on the command that mySQL uses in order to enter data into the table. Why is this not working? My query2 should be successful, but idk why its not.
Please help.
Thanks
to prevent sql injections, try mysqli or pdo
here is mysqli prepared statements version. However if you are trying to create user management system, I wouldn't recommend you do it. There are so many scripts which provide more security, http://www.usercake.com is a good user management system.
session_start();
$db = new mysqli('localhost', 'root', 'password', 'database');
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
if (isset($_POST['username'] && $_POST['pass'] && $_POST['login']))
{
$user_name = ''; //define these here.
$pass = '';
$stmt = $db->prepare("select * from users where username = ? and password = ?");
echo $db->error;//this will echo the error.
$stmt->bind_param('ss', $user_name, $pass);
$stmt->execute();
$result = $stmt->get_result();//get rows
if($result->num_rows < 1) //check if result is less than 1
{
$error = "<br />Wrong Username or Password";}
else{
$_SESSION['user'] = $_POST['username'];
header("Location: home1.php");
}
}
if (isset($_POST['register'])){
$uname = $_POST['newUser'];
$pass = $_POST['newPass'];
if(empty($uname))
{
echo "Please enter your username.";
}
elseif(empty($pass))
{
echo "Please enter your password.";
}
else{
$stmt = $db->prepare("insert into Social.users (username, password) values (?,?)");
echo $db->error;//this will echo the error.
$stmt->bind_param('ss', $uname, $pass);
$stmt->execute();
echo "You have successfully registered.";
}
}
The variables in the INSERT must be the username and password
$sql2 = sprintf("INSERT INTO Social.users (username, password) VALUES (%s, %s)", $_POST['username'], $_POST['pass']);
Use prepared statements and parameterized queries. These are SQL statements that are sent to and parsed by the database server separately from any parameters. This way it is impossible for an attacker to inject malicious SQL.
You basically have two options to achieve this:
Using PDO:
$stmt = $pdo->prepare('SELECT * FROM employees WHERE name = :name');
$stmt->execute(array('name' => $name));
foreach ($stmt as $row) {
// do something with $row
}
Using mysqli:
$stmt = $dbConnection->prepare('SELECT * FROM employees WHERE name =
?'); $stmt->bind_param('s', $name);
$stmt->execute();
$result = $stmt->get_result(); while ($row = $result->fetch_assoc()) {
// do something with $row }
Font: How can I prevent SQL injection in PHP?

PHP Prepared Statement Problems [duplicate]

This question already has answers here:
mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement
(2 answers)
Closed 1 year ago.
slowly getting used to PHP prepared statements, however still get this error
"Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in C:\Users\PC\Documents\XAMPP\htdocs\login.php on line 20".
<?php
$mysqli = new mysqli('localhost', 'c3337015', 'c3337015', 'members');
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if(isset($_GET['loginEmail'])){
session_start();
$stmt = $mysqli->prepare("SELECT Email FROM members WHERE Email=? AND Password=? LIMIT 1");
$email = $_GET['loginEmail'];
$password = $_GET['loginPassword'];
$password = sha1($password);
$stmt->bind_param('ss', $email, $password);
$stmt->execute();
$stmt->bind_result($email, $password);
$stmt->store_result();
if($stmt->num_rows == 1) //To check if the row exists
{
while($stmt->fetch()) //fetching the contents of the row
{$_SESSION['Logged'] = 1;
$_SESSION['Email'] = $email;
header('Location: index.php');
exit();
}
}
else {
echo "Wrong Username or Password!";
}
$stmt->close();
}
else
{
echo "Something went Wrong";
}
$mysqli->close();
?>
$stmt->bind_result($email, $password);
You are binding 2 variables, but only asking for one: SELECT Email FROM members.
I'd also suggest using different variables for bind_result, as it and bind_param both work on references.
$stmt->bind_result($userEmail);

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