How can i convert this statement into mysqli. i tried converting it but is not working. what am i missing. I want to connect to a login form.
This is the code i converted.
<?php
include('admin/dbcon.php');
session_start();
$username = $_POST['username'];
$password = $_POST['password'];
/* teacher */
$query_teacher = mysql_query("SELECT * FROM teacher WHERE username='$username' AND password='$password'")or die(mysql_error());
$num_row_teacher = mysql_num_rows($query_teacher);
$row_teahcer = mysql_fetch_array($query_teacher);
/* admin */
$query_admin = mysql_query("SELECT * FROM users WHERE username='$username' AND password='$password'")or die(mysql_error());
$num_row_admin = mysql_num_rows($query_admin);
$row_admin = mysql_fetch_array($query_admin);
if ($num_row_teacher > 0){
$_SESSION['id']=$row_teahcer['teacher_id'];
echo 'true';
}else if ($num_row_admin > 0){
$_SESSION['id']=$row_admin['user_id'];
echo 'true_admin';
}else{
echo 'false';
}
?>
This is the converted mysqli but i cannot still log on am i missing something. I will be very grateful if you can help me solve this.
<?php
// establishing the MySQLi connection
$con = mysqli_connect("localhost","root","","retreat");
if (mysqli_connect_errno())
{
echo "MySQLi Connection was not established: " . mysqli_connect_error();
}
// checking the user
session_start();
$username = mysqli_real_escape_string($con,$_POST['username'];
$password = mysqli_real_escape_string($con,$_POST['password'];
/* teacher */
$query_teacher = "SELECT * FROM teacher WHERE username='$username' AND password='$password'";
$num_row_teacher = mysqli_query($con,$query_teacher);
$row_teahcer = mysqli_num_rows($num_row_teacher);
/* admin */
$query_admin = "SELECT * FROM users WHERE username='$username' AND password='$password'";
$num_row_admin = mysqli_query($con,$query_admin);
$row_admin = mysqli_num_rows($num_row_admin);
if ($row_teahcer > 0){
//$_SESSION['user_email']=$email;
$_SESSION['']=$row_teacher['teacher_id'];
echo 'true';
}else if ($num_row_admin > 0){
$_SESSION['id']=$row_admin['user_id'];
echo 'true_admin';
}else{
echo 'false';
}
?>
<?php
//db details
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'retreat';
//Connect and select the database
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);
if ($db->connect_error) {
die("Connection failed: " . $db->connect_error);
}
session_start();
$username = $_POST['username'];
$password = $_POST['password'];
/* teacher */
$query = $db->query("SELECT * FROM teacher WHERE username='$username' AND password='$password'")or die(mysqli_error());
$num_row_teacher = mysqli_num_rows($query);
$row_teahcer = mysqli_fetch_array($query);
/* admin */
$query_admin = $db->query("SELECT * FROM users WHERE username='$username' AND password='$password'")or die(mysqli_error());
$num_row_admin = mysqli_num_rows($query_admin);
$row_admin = mysqli_fetch_array($query_admin);
if ($num_row_teacher > 0){
$_SESSION['id']=$row_teahcer['teacher_id'];
echo 'true';
}else if ($num_row_admin > 0){
$_SESSION['id']=$row_admin['user_id'];
echo 'true_admin';
}else{
echo 'false';
}
enter code here
?>
Your are missing
$row_teahcer=mysqli_fetch_array($num_row_teacher);
$row_admin=mysqli_fetch_array( $num_row_admin);
Related
here am trying to get username and password from the database and if their result is found then redirect to some page but mysqli_num_rows returns 0 always i dunno why `
if (isset($_POST['login'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$query = "SELECT * FROM `login` WHERE username = '$username' AND password = '$password'";
$run = mysqli_query($con,$query);
if (mysqli_num_rows($run) == 1) {
header("location: login.php");
}}?>
`
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ornament"; // My local DB Name
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$username = "test"; // $_POST["username"] value
$password = "test"; // $_POST["password"] value
$sql = "SELECT * FROM `ornament` WHERE username = '$username' AND password = '$password'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) == 1) {
echo "correct";
} else {
echo "wrong";
}
mysqli_close($conn);
?>
</body>
</html>
OUTPUT:
correct
I have a database name and table name are same, But for me it's working fine.
NOTE: Please check the whether you are using correct password while getting from POST.
Any how you got a solution, Have a great day
Thanks
Muthu
I was trying to make a page where you can log in and then change your nickname or/and password. Everything in mySQL database, but when I try to save the id to session variable, it doesn't work. Any suggestions?
I am using XAMPP, users is my table in database users, I'm not posting login form code, because it's very simple.
Everything is connected, code doesn't give any warnings or errors.
login.php (fragment):
$sql = "SELECT * FROM users WHERE nickname = '$myusername' and pass = '$mypassword' and confirmed = 1";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$count = mysqli_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count == 1) {
$logged = true;
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"];
$_SESSION['currentId'] = $row["id"];
echo 'Id: ' . $_SESSION['currentId'];
}
}else {
$error = "Your Login Name or Password is invalid";
}
}
change.php (whole):
<?php
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Users";
$currentId = $_SESSION['currentId'];
if($currentId<1){echo 'No Id.';}
else {echo 'CurrentId: ';
echo $currentId;}
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully <br>";
if($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password sent from form
$aCUname = mysqli_real_escape_string($conn,$_POST['CUname']);
$aCUpass = mysqli_real_escape_string($conn,$_POST['CUpass']);
$sql = "UPDATE users SET nickname = '$aCUname', pass = '$aCUpass' WHERE id = '$currentId';";
$result = mysqli_query($conn,$sql);
echo 'Updated successfully.';
}
?>
Thanks for help.
I got a solution. I just had to delete
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
from login.php. Thanks to #Shashikumar Misal !
I am trying to loop through my database and check to see if the user already exists in another table. If they do then I want to increment a value, if they don't then I want to add the user.
When I run the code below it happily loops through all the results:
<?php
$servername = "p:10*********";
$username = "*******";
$password = "*******";
$dbname = "******";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM payroll WHERE user != ' ' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo $result->num_rows;
while($row = $result->fetch_assoc()) {
$user = $row['user'];
$time = $row['time'];
$id = $row['id'];
echo $id;
echo $user;
}
} else {
echo "0 results";
}
$conn->close();
?>
However when I add in the SQL to check to see if they exist in the other table the loop no longer functions correctly and echos the same user each time.
<?php
$servername = "*******";
$username = "******";
$password = "********";
$dbname = "*****";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM payroll WHERE user != ' ' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo $result->num_rows;
while($row = $result->fetch_assoc()) {
$user = $row['user'];
$time = $row['time'];
$id = $row['id'];
echo $id;
echo $user;
// Added existing user check:
$sql = "SELECT * FROM smsreport WHERE user = '$user'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "found";
} else {
echo "USER NOT FOUND";
}
}
} else {
echo "0 results";
}
$conn->close();
?>
In the open eye:
Rename the inside $result variable. It is over writting the first $result.
It could be the problem. Not tested though.
I got little problem with my code... because I try to SELECT sth from database and then INSERT some value to another table, but in normal code from w3school
and I got error
Tryingo to get property of non-object
Here is my code:
<?php
session_start();
function connectionDB(){
$host = "localhost";
$username = "root";
$password = "";
$db_name = "project";
$conn = new mysqli($host, $username, $password, $db_name);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
return $conn;
}
function getID(){
$id = $_GET['id'];
return $id;
}
$login =$_SESSION['login'];
$conn =connectionDB();
$idCar = getID();
echo $idCar;
$sqluser = "SELECT ID_USER FROM login_table WHERE LOGIN = $login";
if($result->num_rows > 0)
$result = $conn->query($sqluser);
{
while($row = $result->fetch_assoc())
{
$sql = "INSERT INTO cart (id_user) VALUES ('".$row['ID_USER']."')";
}
}else echo"error";
?>
THIS IS THE CODE OF SIDE WITH PRODUCT
Please see change near your IF loop
$sqluser = "SELECT ID_USER FROM login_table WHERE LOGIN = $login";
$result = $conn->query($sqluser); //check result first
if($result->num_rows > 0) //get number of rows
{
while($row = $result->fetch_assoc())
{
$sql = "INSERT INTO cart (id_user) VALUES ('".$row['ID_USER']."')";
}
}else echo"error";
Its a simple registration with username and password, but when I try to insert it to the database it always fails and I don't know why, can someone help me with this?
include "db.php";
if (isset($_POST['submit']))
{
$username= $_POST['leguser'] ;
$password= $_POST['legpass'] ;
$pwhash = password_hash($password, PASSWORD_DEFAULT) ;
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);
echo var_dump ($username);
echo var_dump ($password);
echo var_dump ($pwhash);
$sql = "SELECT * FROM tbl_users WHERE fld_username = '$username'";
$result = $conn->query($sql);
if ($result->num_rows === 1) {
echo "<script>alert('Username already used!');</script>"; }
else
{
$q = "INSERT INTO `tbl_users` (`fld_username`) VALUES ('$username')";
$result = mysql_query($q);
if ($result) {
echo 'success';
} else {
echo 'failure';
}
}
This is the code of the database connection (db.php)
<?php
$DB_HOST = 'localhost';
$DB_USER = 'root';
$DB_PASS = '';
$DB_NAME = 'rsi_db';
$conn = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($conn->connect_errno > 0) {
die('Connection failed [' . $conn->connect_error . ']');
}
By the way I only want to input only one data on the database which is the username, for me not to get complicated with every data that I want to put in the future.