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Apr 11, 2006, 11:41 PM
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#1
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DriverHeaven Lover
Join Date: Oct 2005
Location: Behind you.
Posts: 241
Rep Power: 0
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PHP - MySQL: Help (noobie)
I just started learning PHP and MySQL, and I made a code so someone can login with a username and a password. For some reason it's not working, here's the script:
Code:
<?
$mysql_link = mysql_connect("localhost","ANALAUR_kenny","flaco");
@mysql_select_db("ANALAUR_contact", $mysql_link)
if (($username) && ($password)) {
$query = "SELECT user, pass FROM login ";
$query .= "WHERE user='$username' AND pass='$password'";
$result = mysql_query($query, $mysql_link);
if(mysql_num_rows($result)) {
$query = "UPDATE from login SET logged=SYSDATE()";
$query .= "WHERE user='$username' AND pass='$password' ";
mysql_query($query,$mysql_link);
} else {
print("Sorry, this login is invalid.");
exit;
}
} else {
?>
<link rel="stylesheet" href="styleme.css">
<center>
<h1>Main Deck Login </h1>
<table class="leger3" border="2"><tr><td>
<form action="login.php" method="POST">
<table>
<tr><td>Username</td>
<td><input type="text" name="user" size="16"></td></tr>
<tr><td>Password</td><td><input type="password" name="pass" size="16">
</td></tr><tr><td colspan="2"><div align="right">
<input type="submit" value="Login">
</div></td></tr>
</table></form>
</td></tr></table>
</center>
<div align="center"><br>
<span class="style6">Copyright Main Deck Imports, LLC.
</span></div>
<?
}
?>
My database's name is ANALAUR_contact and the table to login is called login. Thanks for the help, and if you have any other SIMPLE login forms you would like to share feel welcome, thanks.
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Apr 13, 2006, 07:49 PM
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#2
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DriverHeaven Lover
Join Date: Oct 2005
Location: Behind you.
Posts: 241
Rep Power: 0
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Well If anyone was interested, I finally got around to it:
Code:
<?php
//Open Database connection
include 'library/db.php';
include 'library/common.php';
dbConnect('ANALAUR_contact');
//Get Variables
$username = $_POST['user'];
$passwordue = $_POST['pass'];
//Encrypt Password
$password = md5($passwordue);
//Set SQL query string
$query = "SELECT `password` FROM `users` WHERE `username` = '$username' LIMIT 1";
//Send String
$result = mysql_query($query);
//Get Results
$row = mysql_fetch_array($result);
//Get DataBase Password to cross-check
$dbpass = $row['password'];
//Check to see if Password entered is the same as password in database
if($password == $dbpass AND $password !== NULL){
session_start();
$_SESSION['user'] = $username;
$_SESSION['pass'] = $password;
?>
<!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Logged in Succesfully</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<link rel="stylesheet" href="library/styleme.css">
</head>
<body>
<p>You have logged in succesfully. <a href="contacts.php">Click here</a> to continue.</p>
</body>
</html>
<?php
} else {
echo("The username and/or password you entered was incorrect."); //Inform User about Login State
}
?>
I'm Still having a problem with this one piece of code, so if anyone can help me:
Code:
<?php
$self = $_SERVER['PHP_SELF'];
if(isset($_GET['id'])) {
$idNum = $_GET['id'];
$query = 'SELECT * FROM `contacts` WHERE `ID` = $idNum';
$result = mysql_query($query) or die('Error, query failed 2');
/* $i = 0;
while($row = mysql_fetch_array($result)) {
echo $row[$i];
++$i;
} */
} else {
$query = 'SELECT * FROM `contacts`';
$result = mysql_query($query) or die('Error, query failed 1');
while($row = mysql_fetch_array($result)) {
$id = $row['ID'];
$contact = $row['Name'] . ' ' . $row['Last Name'] . '<br>';
echo "<a href=\"$self?id=$id\">$contact</a>";
}
}
?>
It's the first part that is not working. I get this "Error, query failed 2" I put a 2 and a 1 so I can recognize where the error was coming from.
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Apr 13, 2006, 07:51 PM
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#3
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DriverHeaven Lover
Join Date: Oct 2005
Location: Behind you.
Posts: 241
Rep Power: 0
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Oh I forgot to put the included documents:
Code:
include 'library/db.php';
include 'library/common.php';
Code:
<?php // db.php
$dbhost = 'mysql.hosting-advantage.com';
$dbuser = 'ANALAUR_kenny';
$dbpass = 'flaco';
function dbConnect($db="") {
global $dbhost, $dbuser, $dbpass;
$dbcnx = @mysql_connect($dbhost, $dbuser, $dbpass)
or die("The site database appears to be down.");
if ($db!="" and !@mysql_select_db($db))
die("The site database is unavailable.");
return $dbcnx;
}
?>
Code:
<?php // common.php
function error($msg) {
?>
<html>
<head>
<script language="JavaScript">
<!--
alert("<?=$msg?>");
history.back();
//-->
</script>
</head>
<body>
</body>
</html>
<?
exit;
}
?>
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Apr 13, 2006, 09:50 PM
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#4
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Tail Razer
Join Date: Jun 2005
Location: Bernyurass, AZ - USA
Posts: 3,757
Rep Power: 28
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you shoudnt post your MYSQL UN/PW - it could comprimise the whole sever , I assume your on a virtual host...
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Apr 14, 2006, 01:10 AM
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#5
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Elite Motherf#$%er
Join Date: Jan 2004
Location: Sacramento, CA
Posts: 2,833
Rep Power: 0
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Is '++$i' possible? I'm pretty sure it's '$i++'...
Also, dunno if this helps or not, but the method I connect to and call things from the db is:
Code:
$connect = mysql_connect(SQL_ADDR, SQL_USER, SQL_PASS);
mysql_select_db(SQL_DBASE, $connect);
$sql = mysql_query("SELECT * FROM news WHERE newsActive='1' ORDER BY newsDate DESC");
while($row = mysql_fetch_array($sql)) {
...
}
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Apr 14, 2006, 02:15 PM
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#6
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DriverHeaven Lover
Join Date: Oct 2005
Location: Behind you.
Posts: 241
Rep Power: 0
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Well, the password and username I posted there were incorrect, but the host is what I use.
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