What is wrong with this code? Why can't I get the username from the members table in website database. Here's the screenshot of database information:
I am trying to make a function for it so that I can use it later. Here's the code:
<?php
function get_username($username){
$host = 'localhost';
$username = 'root';
$password = '';
$database = 'website';
$con = mysqli_connect($host, $username, $password, $database);
$query = "SELECT * FROM members WHERE username = '{$username}'";
$result = mysqli_query($con, $query);
$row = mysqli_fetch_array($result);
return $row['name'];
mysqli_close($con);
}
echo get_username('iamfaizahmed');
?>
Your username is overwritten in the function
$username = 'root'; // here
So no matter what you pass in the function argument it will always use
where username = 'root'
So use a different variable name
you are mixing between $username of the function and $username of your connect variable
change this
function get_username($username){
to
function get_username($user){
and do your query like that
$query = "SELECT * FROM members WHERE username = '$user'";
First thing, your $username variable.
You're passing it through the function and also using it for db connection.
Change your dbusername in the function as $dbusername
Second, the query. Keep it as:
$query = "SELECT * FROM members WHERE username = '$username'";
Now, once you're done with the basics, I recommend you to visit this link: Click Here
That should get your somewhere!
Related
I have this code:
<?php
$user = $_COOKIE["user"];
$password = $_COOKIE["password"];
$localhost = "localhost";
$userdb = "xxxxx";
$passworddb = "xxxxx";
$database = "xxxxx";
$conn = mysqli_connect($localhost, $userdb, $passworddb, $database);
$vyber = "SELECT PASSWORD FROM Login WHERE User=".$user;
$result = mysqli_query($conn, $vyber);
echo $result;
?>
Cookie are set and if I use $vyber in database so everything is good. But there PHP write nothing. Can anybody tell, what I doing wrong? (Without comand $vyber every thing running perfect)
instead of,
echo $result
try to do that :
while ($row = mysqli_fetch_row($result)){
echo $row[0];
}
It is query error change query to :
$vyber = "SELECT PASSWORD FROM Login WHERE User='$user'";
if did not work use die function to dispaly error message :
mysqli_query($conn, $vyber) or die(mysqli_error($conn));
To fetch records :
while ($row = mysqli_fetch_array($result)){
echo $row[0];
}
I am connected to the database, a page has lots of content so I'll only share the part that doesnt return a value in php, but it returns a value in MySQL
Here is the code;
$query = "SELECT firstname FROM users WHERE id = '17'";
$query_run = mysql_query($query);
$row = mysql_fetch_row($query_run);
echo $row[0];
Changing the code I shared first to this, solved the problem. Thanks to anyone who tried to help.
$query = "SELECT firstname FROM users WHERE id = '17'";
$query_run = mysqli_query($conn, $query);
$row = mysqli_fetch_row($query_run);
echo $row[0];
And made bit changes to the connect.inc.php which I also shared in comment.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "notsitesi";
$conn = new mysqli($servername, $username, $password, $dbname);
if($conn->connect_error) {
die("Connection failed");
}
?>
I'm getting an error saying Undefined variable: con,
the connection of the database is on the other php file, (include() is already on top of the code). I just don't know how to call the $con
if (isset($_POST['update_profile']))
{
if (isset($_POST['first_name']))
{
$first_name = mysqli_real_escape_string($con, $_POST['first_name']);
$sql = mysqli_query($con, "UPDATE tbl_fbusers SET fname = '$first_name' WHERE email = '$email_to_connect'");
}
if (isset($_POST['last_name']))
{
$last_name = mysqli_real_escape_string($con, $_POST['last_name']);
$sql = mysqli_query($con, "UPDATE tbl_fbusers SET lname = '$last_name' WHERE email = '$email_to_connect'");
}
if (isset($_POST['contact']))
{
$contact = mysqli_real_escape_string($con, $_POST['contact']);
$sql = mysqli_query($con, "UPDATE tbl_fbusers SET contact = '$contact' WHERE email = '$email_to_connect'");
}
}
here is the other php file
class Users {
public $table_name = 'tbl_fbusers';
function __construct(){
//database configuration
$dbServer = 'localhost'; //Define database server host
$dbUsername = 'root'; //Define database username
$dbPassword = ''; //Define database password
$dbName = 'db_zalian'; //Define database name
//connect databse
$con = mysqli_connect($dbServer,$dbUsername,$dbPassword,$dbName);
if(mysqli_connect_errno()){
die("Failed to connect with MySQL: ".mysqli_connect_error());
}else{
$this->connect = $con;
}
}
Thanks!
Defining $con as a GLOBAL variable is a terrible idea...
I suggest to make a file (eg. connection.php) that will contain the $con variable that is not in a function, and then include the connection.php to your other php files. It's more secure and easier, and you won't get to any troubles.
Since you have a class you need to initialize user class.
$user = new Users();
and then
$con = $user->connect;
Here you can run your sql like:
$contact = mysqli_real_escape_string($con, $_POST['contact']);
$sql = mysqli_query($con, "UPDATE tbl_fbusers SET contact =......... etc.
you need to define $con as global:
global $con
$con = mysqli_connect($dbServer,$dbUsername,$dbPassword,$dbName);
I'm trying to build a login process where, by using $_SESSION variables, the login credentials of the user are stored and used to show their relevant data from the database on screen (i.e. they will only see the school data that they work for).
<?php
session_start();
if(!isset($_SESSION['Initials'], $_SESSION['Surname']))
{
$host = "xxx";
$username = "xxx";
$password = "xxx";
$database_name = "xxx";
$table_name = "xxx";
mysql_connect($host, $username, $password) OR die("Can't
connect");
mysql_select_db($database_name) OR die("Can't connect to
Database");
$query = "SELECT Class FROM $table_name WHERE Initials = '".
$_SESSION['Initials']."' AND staff LIKE '%".$_SESSION['Surname']."'";
$result = mysql_query($query);
$class = mysql_fetch_array($result);
$count = mysql_num_rows($result);
if($count === NULL)
{
echo "ERROR";
}
else
{
$_SESSION['Class'] = $result;
echo "Class added to sessions";
}
}
?>
My initial problem where the query couldn't recognize the session variables was easily solved by adding the correct brackets for the if-statement. My next problem that has arisen here is that even though the query should be successfull (I don't receive an error message saying 'ERROR' when the $count is either FALSE or NULL) it's not creating the result array into a new session, because when I print the session array on a new page it's still only carrying over the 'Initials' and 'Surname' sessions.
What do I need to change to my query, or post-query process in order for that array (because it's bound to throw up multiple results) to be made into a new session?
Many thanks for the answers to my initial problem!
if(!isset($_SESSION['Initials'], $_SESSION['Surname'])) {
// code
}
u need { } brackets
if(!isset($_SESSION['Initials'], $_SESSION['Surname']))
$host = "xxxxx"; $username = "xxxxx"; $password = "xxxxx";
is
if(!isset($_SESSION['Initials'], $_SESSION['Surname'])) {
$host = "xxxxx";
}
$username = "xxxxx";
$password = "xxxxx";
I've found the answer - it turned out that I wasn't treating one of the session variables as a proper array and thus wouldn't load properly. I've added my script below so that people with similar problems in the future can use it as a reference point:
<?php
session_start();
// Server Details //
$host = "---";
$username = "---";
$password = "---";
$database_name = "---";
$table_name = "---";
// Connect Command //
mysql_connect($host, $username, $password) OR die("Can't
connect");
mysql_select_db($database_name) OR die("Can't connect to
Database");
// Query to call up the unique school name //
$query_school = mysql_query("SELECT DISTINCT School FROM $table_name
WHERE Initials = '".$_SESSION['---']."'
AND staff LIKE '%".$_SESSION['---']."'") or die( mysql_error());
$result_school = mysql_result($query_school, 0);
// Query to call up the unique centre no //
$query_centreno = mysql_query("SELECT DISTINCT CentreNo FROM
$table_name WHERE Initials = '".$_SESSION['---']."'
AND staff LIKE '%".$_SESSION['---']."'") or die( mysql_error());
$result_centreno = mysql_result($query_centreno, 0);
// The newly created sessions for school info //
$_SESSION['---'] = $result_school;
$_SESSION['---'] = $result_centreno;
// Query to call up the array of classes //
$query_class = mysql_query("SELECT Class FROM $table_name WHERE
Initials = '".$_SESSION['---']."'
AND staff LIKE '%".$_SESSION['---']."'") or die( mysql_error());
$query_class__array = array();
while($row = mysql_fetch_assoc($query_class))
$query_class_array[] = $row;
$_SESSION['---'] = $query_class_array;
?>
I'm new to PHP and SQL but I'm trying to create a simple PHP-script that allows a user to login to a website. It doesn't work for some reason and I can't see why. Every time I try to login with the correct username & password, I get the error "Wrong Username or Password". The database-name and table-name are correct.
connect.php:
<?php
$db_host = 'localhost';
$db_name = 'app';
$db_user = 'root';
$db_pass = '';
$tbl_name = 'users';
// Connect to server and database
mysql_connect("$db_host", "$db_user", "$db_pass") or die("Unable to connect to MySQL.");
mysql_select_db($db_name)or die("Cannot select database.");
// Info sent from form
$user = trim($_POST['user']);
$pass = trim($_POST['pass']);
// Protection against MySQL injection
$user = stripslashes($user);
$pass = stripslashes($pass);
$user = mysql_real_escape_string($user);
$pass = mysql_real_escape_string($pass);
$sql = ("SELECT * FROM $tbl_name WHERE username='$user' and password='$pass'");
$result= mysql_query($sql);
$count 0= mysql_num_rows($result);
if($count==1){
// Register $user, $pass send the user to "score.php"
session_register("user");
session_register("pass");
header("location:score.php");
}
else
{
echo "Wrong Username or Password";
}
?>
score.php:
<?php
session_start();
if(!session_is_registered(user)){
header("location:login.html");
}
?>
<html>
<body>
<h1>Login Successful</h1>
</body>
</html>
I hope someone can find my mistake, thanks!
FYI session_register and session_is_registered are deprecated and will be removed from PHP. Also try to change your code to use mysqli or PDO. Plenty of articles explain how to do it. Finally, make sure you escape input from the user ($_POST array) because you never know what the user will send and you don't want to be prone to SQL injections. You really do not want to store passwords in clear text, so using SHA1 or MD5 is best.
Having written the above, your code becomes (you can use the $_SESSION global array directly):
connect.php:
<?php
$db_host = 'localhost';
$db_name = 'app';
$db_user = 'root';
$db_pass = '';
$tbl_name = 'users';
// Connect to server and database
mysql_connect($db_host, $db_user, $db_pass) or die("Unable to connect to MySQL.");
mysql_select_db($db_name) or die("Cannot select database.");
// Info sent from form
$user = trim($_POST['user']);
$pass = trim($_POST['pass']);
// Protection against MySQL injection
$user = stripslashes($user);
$pass = stripslashes($pass);
$user = mysql_real_escape_string($user);
$pass = mysql_real_escape_string($pass);
$sql = "SELECT * FROM $tbl_name "
. "WHERE username = '$user' "
. "AND password = sha1('$pass')";
$result = mysql_query($sql);
// There was an extra 0 here before the equals
$count = mysql_num_rows($result);
if ($count==1)
{
// Register $user, $pass send the user to "score.php"
$_SESSION['user'] = $user;
// You really don't need to store the password unless you use
// it somewhere else
$_SESSION['pass'] = $pass;
header("location: ./score.php");
}
else
{
echo "Wrong Username or Password";
}
?>
score.php:
<?php
session_start();
if (!isset($_SESSION['user']))
{
header("location:login.html");
}
?>
<html>
<body>
<h1>Login Successful</h1>
</body>
</html>
A couple of things
Change this line to the one with error checking i have put below it
$result= mysql_query($sql);
$result= mysql_query($sql) or die(mysql_error());
chances are there is an sql error and you are not picking it up, so the result will always have 0 rows
Also not sure if this line is a typo or not, there shouldn't be a 0 in there
$count 0= mysql_num_rows($result);