Log-In Box

[Salchester]

Hiya,

I'm new to PHP, how do I make a simple Login page that compares the Username and Password entered in a form, with whats in a table.

If login is unsuccessful, how do I display an error message.

If login is successful, how do I navigate to a different page, and keep the user logged-in until the user chooses to logout?

[Dan Grossman]

session_start();
mysql_connect("localhost", "username", "password");
mysql_select_db("some_db");

$username = $_POST['username'];
$password = $_POST['password'];

$sql = "SELECT id FROM users WHERE username = '" . mysql_real_escape_string($username) . "' AND password = '" . mysql_real_escape_string($password) . "'";
$result = mysql_query($sql);
if (mysql_num_rows($result) == 0) {
  echo "Username and password combination not found.";
} else {
  $row = mysql_fetch_array($result);
  $_SESSION['logged_in'] = 1;
  $_SESSION['user_id'] = $row['id'];
  header("Location: http://www.example.com/member-area.php");
} 

On your pages that require authorization, check for the presence of those session values:

session_start();
if (!isset($_SESSION['logged_in'])) {
  //Not logged in, redirect to login page
  header("http://www.example.com/login.php");
}
//If they pass that, they're logged in, and are authorized to view whatever else is in this file 

[Salchester]

How do I displayed the users Username on every page once the user has logged in, along with a "Logout" link, that upon being clicked logs the user out of the system resulting in the user having to log back in again to return to the members area?

[Dan Grossman]

Add the username to the query and session array in the login code, then echo that variable on your other pages. The logout page needs just this code:

session_start();
unset($_SESSION['logged_in']); 

[Salchester]

Add the username to the query and session array in the login code, then echo that variable on your other pages.

How do I do that, as I'm new to PHP & sessions?

So the logout page only needs the following code? Doesn't the session need to be destroyed to logout, in order for the user to have to re-login?

session_start();
unset($_SESSION['logged_in']);

[Salchester]

Upon running the following code, I get greeted with the following error message:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in
C:\Documents and Settings\It's Dale\Desktop\xampplite\htdocs\Login.php on line 21

<?php

session_start();

mysql_connect("localhost", "peter", "abc123");

mysql_select_db("Users");



$Username = $_POST['Username'];

$Password = $_POST['Password'];



$sql = "SELECT id FROM users WHERE Username = '" . mysql_real_escape_string($username) . "' AND password = '" . mysql_real_escape_string($password) . "'";

$result = mysql_query($sql);

if (mysql_num_rows($result) == 0) {

  echo "Username and password combination not found.";

} else {

  $row = mysql_fetch_array($result);

  $_SESSION['logged_in'] = 1;

  $_SESSION['user_id'] = $row['id'];

  header("Location: http://www.example.com/member-area.php");

} 
?>

[william232]

dont u need to use the mysql connect link which is $con or watever

then just replace it cuz looks like its not connectingg to database

[Salchester]

dont u need to use the mysql connect link which is $con or watever

then just replace it cuz looks like its not connectingg to database

What do you mean? Replace what?

[william232]

so its like

$con=mysql_connect("localhost", "peter", "abc123");

mysql_select_db("Users")

then on querys have it

$result = mysql_query($sql,$con);

Try that.

[Dan Grossman]

so its like

$con=mysql_connect("localhost", "peter", "abc123");

mysql_select_db("Users")

then on querys have it

$result = mysql_query($sql,$con);

Try that.

mysql_query() will use the last opened link if that parameter is omitted.

[Salchester]

I'm new to PHP, how do I display the word "Log-Out" on the screen, which upon being clicked, logs the user out of the system. Requiring the user to re-login to review the members page?

[william232]

that is the way i do it dan