PHP - Trouble With Login Scripting Using Php & Mysql
This is my first attemp at a log in system for a website. Everything seems to work fine until the "successful" IF function near the end. All I get it an output of "?>" instead of a redirect to the file "login_success.php". Any help would be GREATLY appreciated!! Tom
<?php // Connect to server and select databse. mysql_connect("localhost", "scripts3_public", "sfj123!")or die("cannot connect"); mysql_select_db("scripts3_sfj")or die("cannot select DB"); // username and password sent from form $fusername=$_POST['fusername']; $fpassword=$_POST['fpassword']; // To protect MySQL injection (more detail about MySQL injection) $fusername = stripslashes($fusername); $fpassword = stripslashes($fpassword); $fusername = mysql_real_escape_string($fusername); $fpassword = mysql_real_escape_string($fpassword); $sql="SELECT * FROM `users` WHERE `User name` = '$fusername' AND `Password` = '$fpassword'"; $result=mysql_query($sql); if(!mysql_num_rows($result)) {echo "No results returned.";} // Mysql_num_row is counting table row $count=mysql_num_rows($result); // If result matched $fusername and $fpassword, table row must be 1 row if($count==1){ // Register $fusername, $fpassword and redirect to file "login_success.php" session_register("fusername"); session_register("fpassword"); header("location:login_success.php"); } else { echo "Wrong Username or Password"; } ?> Similar Tutorialserrors: Deprecated: Function session_register() is deprecated in /Applications/XAMPP/xamppfiles/htdocs/login.php on line 18 Warning: session_register() [function.session-register]: Cannot send session cookie - headers already sent by (output started at /Applications/XAMPP/xamppfiles/htdocs/login.php:18) in /Applications/XAMPP/xamppfiles/htdocs/login.php on line 18 Warning: session_register() [function.session-register]: Cannot send session cache limiter - headers already sent (output started at /Applications/XAMPP/xamppfiles/htdocs/login.php:18) in /Applications/XAMPP/xamppfiles/htdocs/login.php on line 18 Deprecated: Function session_register() is deprecated in /Applications/XAMPP/xamppfiles/htdocs/login.php on line 22 Code: Code: [Select] <?php if ($_POST['email']) { include_once "connect_to_mysql.php"; $email = stripslashes($_POST['email']); $email = strip_tags($email); $email = mysql_real_escape_string($email); $password = preg_replace("[^A-Za-z0-9]", "", $_POST['password']); $password = md5($password); $sql = mysql_query("SELECT * FROM members WHERE email='$email' AND password='$password' AND emailactivated='1'"); $login_check = mysql_num_rows($sql); if($login_check > 0){ while($row = mysql_fetch_array($sql)){ $id = $row["id"]; session_register('id'); $_SESSION['id'] = $id; $username = $row["username"]; session_register('username'); $_SESSION['username'] = $username; mysql_query("UPDATE members SET lastlogin=now() WHERE id='$id'"); header("location: member_profile.php?id=$id"); exit(); } } else { print '<br /><br /><font color="#FF0000">No match in our records, try again </font><br /> <br /><a href="login.php">Click here</a> to go back to the login page.'; exit(); } } ?> any help really appreciated...thanks!! I am trying to create a login menu, but I keep getting the same errors. Ok here is what I'm trying to do. There is a link in a email that is sent out to the people who use this app. When they click on the email link they are brought to this site which is password protected. So they have to enter their username and password. What I want to below script to do is to log them in while checking to see if the $id variable is set. If it is, then the script is to take them to the page in the email link. If not take them to the submit job page. <?php $id=$_POST["id"]; $cmd = $_POST['cmd']; $connection = mysql_connect("host", "user", "pass"); mysql_select_db("database", $connection) or die(mysql_error()); switch($cmd) { case "login": $u = $_POST['username']; $p = $_POST['password']; $query = "SELECT * FROM login WHERE username='$u' AND password='$p'"; $result = mysql_query($query); $row = mysql_fetch_array($result); if (isset($id))($row){ session_start(); $_SESSION['user_id'] = $row[0]; $_SESSION['residentname'] = $row[1]; $_SESSION['unit_num'] = $row[2]; setcookie("TestCookie", time()+3600); /* expire in 1 hour */ $resite = "submitjob.php?do=viewone&id=$id"; header("Location:$resite"); exit(); } else if ($row){ session_start(); $_SESSION['user_id'] = $row[0]; $_SESSION['residentname'] = $row[1]; $_SESSION['unit_num'] = $row[2]; setcookie("TestCookie", time()+3600); /* expire in 1 hour */ $resite = "submitjob.php"; header("Location:$resite"); exit(); } else { echo "Sorry the app didn't find a match."; } break; } ?> I'm trying to implement sessions into my website. At the moment index.php contains a login form that posts to AccountManagement.php. AccountManagement.php then checks the database to see if they have entered a correct username/password combination. This all works fine, however I would like the site to remember that a user has logged in, and not tell them that they have entered an invalid password every time they come to this page by any means other than index.php's login form (e.g. a back button on a page that follows from AccountManagement). I have tried for days to get this to work using a for loop that checks if the session is started, but I can't seem to get the placement/syntax correct. Any help would be greatly appreciated. AccountManagement.php: Code: [Select] <?php include ("Includes/database.php"); include ("Includes/htmlheader.php"); dbconnect ("localhost", "xxxxx", "xxxxx", "xxxxx"); $query=sprintf("SELECT wowUsername, Password, UserID FROM Users WHERE (((wowUsername)=\"%s\") AND ((Password)=\"%s\"));", $_POST['Username'], $_POST['Password']); $result=mysql_query($query); if (!$result) { $message = 'Invalid query: ' . mysql_error() . "\n"; $message .= 'Whole query: ' . $query; die($message);} if (mysql_num_rows($result) !=1) { $errormessage= "Incorrect Username or Password, please try again."; include ("Includes/error.php"); } else { $row=mysql_fetch_assoc($result); $CustomerID = $row['UserID']; $query2=sprintf("SELECT CustomerID, FName FROM Customers WHERE CustomerID=$CustomerID"); $result2=mysql_query($query2); $row2=mysql_fetch_assoc($result2); $_SESSION['UserID']=$CustomerID; ?> <form action="index.php" id="home" name="home" style="width: 8em"></form> <h1> Account Management </h1> <p><h3 align="center">Welcome <?php echo $row2['FName'];?>, use the buttons below to manage your subscriptions.<h3><br /> <h2> <form action="Subscription.php" id="subs" name="subs"> <p> <input class="button5" name="Setup" type="submit" value="New Subscription" align="center" /></p> </form></h2> <form action="AccountUpdate.php" id="remove" name="remove" style="width: 8em"> <p> <input class="button5" name="NewDetails" type="submit" value="Update Details" /> </p></form> </p> <p> <form action="AccountCancel.php" id="remove" name="remove" style="width: 8em"> <input name="Logout3" type="submit" class="button5" value="Cancel Account" align="right" /> </form> </p> <p> <br /> <form action="index.php" id="remove" name="remove" style="width: 8em"> <input class="button5" name="Logout" type="submit" value="Log Out" /> </p> </p> <?php } ?> </div> </body> </html> </form> htmlheader.php: Code: [Select] <?php error_reporting(E_ERROR | E_WARNING | E_PARSE ); if(!isset($_SESSION)) { session_start(); $_SESSION['UserID']=0; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head><link rel="stylesheet" type="text/css" href="CSS/Styles.css"/> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Account Management</title> </head> <body> </form> <div id="content"> I'm not sure where the issue really lies after the form submits it DOES perform the error messages if there is one, however if the username and password are atleast filled in and the user clicks Log In it doesn't do anything after that. login.php <?php /** * @author Jeff Davidson * @copyright 2010 */ if (isset($_POST['submitted'])) { require_once ('inc/login_functions.php'); require_once ('inc/dbconfig.php'); list ($check, $data) = check_login($dbc, $_POST['username'], $_POST['password']); if ($check) { // OK! // Set the session data:. session_start(); $_SESSION['id'] = $data['id']; $_SESSION['firstname'] = $data['firstname']; // Redirect: $url = absolute_url ('loggedin.php'); header("Location: $url"); exit(); }else { // Unsuccessful! $errors = $data; } mysqli_close($dbc); } // End of the main submit conditional. include ('inc/login_page.php') ?> login_functions.php <?php /** * @author Jeff Davidson * @copyright 2010 */ // This page defines two functions used by the login/logout process. /* This function determines and returns an absolute URL. * It takes one argument: the page that concludes the URL. * The argument defaults to index.php. */ function absolute_url($page = 'index.php') { // Start defining the URL... // URL is http://plus the host name plus the current directory: $url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['PHP_SELF']); // Remove any trailing slashing: $url = rtrim($url, '/\\'); // Add the page $url .= '/' . $page; // Return the URL: return $url; } // End of absolute_url() function. /* This function validates the form data (the username and password). * If both are present, teh database is queried. * The function requires a database connection. * The function returns an array of information, including: * - a TRUE/FALSE variable indicating success * - an array of either errors or the database result */ function check_login($dbc, $username = '', $password = '') { $errors = array(); // Initialize error array. // Validate the username if (empty($username)) { $errors[] = 'You forgot to enter your username.'; } else { $u = mysqli_real_escape_string($dbc, trim($username)); } // Validate the password: if (empty($password)) { $errors[] = 'You forgot to enter your password.'; } else { $p = mysqli_real_escape_string($dbc, trim($password)); } if (empty($errors)) { // If everythings OK. // Retrieve the firstname and lastname for the username/password combination: $q = "SELECT id, firstname FROM users WHERE username='$u' AND password=SHA('$p')"; $r = @mysqli_query($dbc, $q); // Run teh query. // Check the result: if (mysqli_num_rows($r) == 1) { // Fetch the record: $row = mysqli_fetch_array($r, MYSQLI_ASSOC); // Return true and the record: return array(true, $row); }else { // Not a match! $errrors[] = 'The username and password entered do not match those on file.'; } } // End of empty ($errrors) IF. // Return false and the errors: return array(false, $errors); } //End of check_login() function. ?> login_page.php <?php /** * @author Jeff Davidson * @copyright 2010 */ // This page prints any errors associated with logging in and creates the login, including the form. // Prints any error messages, if they exists: if (!empty($errors)) { echo '<h1>Error!</h1> <p class="error">The following error(s) occured:<br />'; foreach ($errors as $msg) { echo " - $msg<br />\n"; } echo '</p><p>Please try again.</p>'; } // Display the form: ?> <!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> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <meta name="description" content="Caracole" /> <title>Titanium</title> <link HREF="favicon.ico" type="image/x-icon" rel="icon" /> <link HREF="favicon.ico" type="image/x-icon" rel="shortcut icon" /> <link rel="stylesheet" type="text/css" href="css/tripoli.simple.css" media="screen, projection, print" /> <link rel="stylesheet" type="text/css" href="css/base.css" media="screen, projection, print" /> <link rel="stylesheet" type="text/css" href="css/layout.css" media="screen, projection, print" /> <link rel="stylesheet" type="text/css" href="css/style.css" media="screen, projection, print" /> <link rel="stylesheet" type="text/css" href="css/theme.css" media="screen, projection, print" /> <link rel="stylesheet" type="text/css" href="css/icons.css" media="screen, projection, print" /> <script type="text/javascript" SRC="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script> <script type="text/javascript"> //<![CDATA[ document.write('<link rel="stylesheet" type="text/css" href="css/js/js.css" media="screen, projection, print" />'); //]]> $(document).ready(function(){ $(".close").click(function(){ $(this).parents(".message").hide("puff"); }); }); </script> <!--[if IE]> <link rel="stylesheet" type="text/css" href="css/ie/ie.css" media="screen, projection, print" /> <![endif]--> <!--[if lt IE 7]> <script src="js/DD_belatedPNG_0.0.7a-min.js" type="text/javascript"></script> <script> DD_belatedPNG.fix(' #header, h1, h1 a, .close, .field,.paginate .current, .icon, .required-icon'); </script> <link rel="stylesheet" href="css/ie/ie6.css" type="text/css" media="screen, projection"/> <![endif]--> </head> <body> <!-- Content --> <div id="login" class="content"> <div class="roundedBorders login-box"> <!-- Title --> <div id="title" class="b2"> <h2>Log In</h2> <!-- TitleActions --> <div id="titleActions"> <div class="actionBlock"> <a href="#">Forgot your password ?</a> </div> </div> <!-- /TitleActions --> </div> <!-- Title --> <!-- Inner Content --> <div id="innerContent"> <form action="login.php" method="post"> <div class="field"> <label for="username">Username</label> <input type="text" class="text" id="username" name="username" /> </div> <div class="field"> <label for="password">Password</label> <input type="password" class="text" id="password" name="password"/> </div> <div class="clearfix login-submit"> <span class="fleft"> <input type="checkbox" name="remember-me" id="remember-me" /> <label for="remember-me">Remember me</label> </span> <span class="fright"> <button class="button" type="submit" name="submit"><strong>Log In</strong></button> </span> </div> <input type="hidden" value="TRUE" name="submitted" /> </form> </div> <!-- /Inner Content --> <div class="bBottom"><div></div></div> </div> </div> </body> </html> loggedin.php <?php /** * @author Jeff Davidson * @copyright 2010 */ // The user is redirected here from login.php. session_start(); // Star the session. // If no session value is present, redirect the user: if (!isset($_SESSION['id'])) { require_once('inc/login_functions.php'); $url = absolute_url(); header("Location: $url"); exit(); } $page_title = 'Logged In!'; // Print a customized message: echo "<h1>Logged In!</h1> <p>You are now logged in, {$_SESSION['firstname']}!</p> <p><a href=\"logout.php\">Logout</a></p>"; ?> I thought I'd come back in and insert the file manager I have setup here. root/loggedin.php root/login.php root/inc/login_page.php root/inc/login_functions.php I am having a little bit of trouble with this piece of code. I'm sure it's something simple, but I have been working on this thing all day and want to get it finally finished. Here's the troublesome code: function rrmdir($dir) { if (is_dir($dir)) { $objects = scandir($dir); foreach ($objects as $object) { if ($object != "." && $object != "..") { if (filetype($dir."/".$object) == "dir") rrmdir($dir."/".$object); else unlink($dir."/".$object); } } reset($objects); rmdir($dir); } } $sql_clean = "SELECT * complete WHERE createdate < date_sub(current_date, interval 1 minute)"; $sql_list = mysql_query($sql_clean); while($row = mysql_fetch_assoc($sql_list)) { $directory = "complete/" . $row['fileurl']; rrmdir($directory); } The purpose of this particular bit is to run on a cron every few days. It gets "createdate" and other info from the "complete" table in order to know how old the record is. If the record is older than (in the example, 1 minute; it will be set to several days on public) the defined max age, it removes that directory and everything within it to keep the directory clean and the disk usage down. The error returned is Quote Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home1/latenit2/public_html/kindleprocessor/process/garbagecleaner.php on line 46 Line 46 is Quote while($row = mysql_fetch_assoc($sql_list)) { I may be doing the look-up on the MySQL database incorrectly, too. I haven't discounted that, and I'd be thankful if someone could help me out with this issue. Can someone please explain to me why I cant seem to get my mysql update line to work. I have been trying for a while an still nothing. I am new in php and need some help getting this to work. Please be gentle. a good explaination in newbie talk would be appreciated. The session variable I echoed out does work so I know I am reading the variable in from the other page. thanks <?php session_start(); /* Server side scripting with php CISS 225 Lab # Final Project */ //This section will create variables collected from information sent //by the post method on the createUserProcess. /* $_SESSION['city'] = $_POST['city']; $_SESSION['state'] = $_POST['state']; $_SESSION['zipCode'] = $_POST['zipCode']; $_SESSION['profession'] = $_POST['profession']; $_SESSION['activities'] = $_POST['activities']; $_SESSION['hobbies'] = $_POST['hobbies']; */ $city = $_POST['city']; $state = $_POST['state']; $zipCode = $_POST['zipCode']; $profession = $_POST['profession']; $activities = $_POST['activities']; $hobbies = $_POST['hobbies']; $db = mysql_connect("localhost", "root", ""); mysql_select_db("accountprofile",$db); echo $_SESSION['Email']; //$query = "UPDATE accountprofile SET city = '$city', state = '$state', zipcode = '$zipCode', profession = '$profession', " . " //activities = '$activities', hobbies = '$hobbies' WHERE lastName = 'Hildebrand'"; $query = "UPDATE accountprofile SET city = '$city', state = '$state', zipcode = '$zipCode', profession = '$profession', activities = '$activities', hobbies = '$hobbies' WHERE userName = " .$_SESSION['Email'].""; mysql_query($query,$db); if (mysql_error()) { echo "$query<br />"; echo mysql_error(); } echo "THANK YOU!<br />"; echo "Your profile has been completed!<br />"; ?> Hey Everybody, I am writing a SUPER SIMPLE script and for some reason I cannot figure this issue out. I guess I'm too close to the situation and have spent too many hours staring at this script. Here's the problem: I am running a basic SQL query through php that should return multiple rows of data and instead returns the first row multiple times. I'm not sure what the problem is, but I'm sure YOU can help! <?php //Get Invoice Rows $sql = 'SELECT * FROM timecard WHERE INVOICE_ID=\'1000\''; $result = mysql_query($sql); $rows = mysql_fetch_array($result); $num = mysql_num_rows($result); //Build Current Invoice $i=0; $invoice = '<table class="invoice" cellspacing="0" cellpadding="0">'; $invoice .= '<tr class="heading"><td>#</td><td>Invoice</td><td>Date</td><td>Time In</td><td>Time Out</td><td>Hours</td><td>$/Hr</td><td>Sub Total</td></tr>'; while($i < $num){ if( $i%2 ) { $eo = 'odd'; } else { $eo = 'even'; } $invoice .= '<tr id="invoiceRow" class="'.$eo.'"><td>'.$rows[0].'</td><td>PHG'.$rows[1].'</td><td>'.$rows[2].', '.$rows[4].' '.$rows[3].', '.$rows[5].'</td><td>'.$rows[6].'</td><td>'.$rows[7].'</td><td>'.$rows[8].'</td><td>'.$rows[9].'</td><td>'.$rows[10].'</td></tr>'; $runningTotal[$i] = $rows[10]; $i++; } //Get Total $total = array_sum($runningTotal); $invoice .= '<tr><td colspan="7" style="background-color: #000000; color: #ffffff; font-weight: bold; padding-left: 5px;">Total</td><td align="right" style="background-color: #333333; font-weight: bold; color: #FFFFFF; padding-right: 5px;">'.$total.'</td>'; $invoice .= '</table>'; echo $invoice; ?> Much thanks in advance for anyone that is able to resolve this problem, even just a try is nice!! Thank You, E This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=313679.0 Hello. Many thanks for your help. I am writing a PHP/MySQL dating-site and have hit a programming impass. I have a database full of members and a search form consisting of checkboxes. So to search, a member ticks say...gender: female; age: 21,22,23,24,25,26; height: 5'4",5'5",5'6",5'7"; county: cornwall,devon,somerset How can a run a check on the database selecting all entries that fall into the selected criteria. For example a 23 year old female of 5'5" living in Cornwall and a 26 year old female of 5'4" living in Somerset? The key index of my database is 'id' and the fields a age,height,county The names of the form checkboxes a Gender: male, female; Age: 21,22,23,24 etc; Height: 5_4,5_5,5_6 etc; county: cornwall,devon etc I am trying to insert values stored within two dimensional array into mysql database but it does not work as I would expect it. The locations in mysql are defined as char length of 2. When I print_r the array it shows: Array( [0] => Array ( [0] => 04 [1] => 22 [2] => 27 [3] => 28 [4] => 39 [5] => 43 [6] => 47 )) but when I insert them into the mysql like this: Number1A='$MaxMillionsNumber[0][0]', Number1B='$MaxMillionsNumber[0][1]', Number1C='$MaxMillionsNumber[0][2]', Number1D='$MaxMillionsNumber[0][3]', Number1E='$MaxMillionsNumber[0][4]', Number1F='$MaxMillionsNumber[0][5]', Number1G='$MaxMillionsNumber[0][6]'; my values in mysql all show as Ar What am I doing wrong? I am using a login system in php and mySQL but only one page is potected. pages i am using: 1. login.php // inputing details (user name, password) 2. checkloginDetails.php // connect to db and check login details 3. logged_in.php // successfully login ...i need more than the one page protected for example; once the user has logged in there will be the main logged in page with other links, remove topics, add, user, remove user all these pages i want protecting but with out the user inputing his details again. Has anyone got an idear onhow i ould achive this? I have created a PHP & MySql login but its not working. If I put the right email/password still its showing "Wrong Username or Password" everytime. Bacause I'm beginner to this I don't really know how to solve this issue. Thanks in advance. Here is my coding <?php // Start PHP session at the beginning session_start(); // Create database connection using config file include_once("connection.php"); // If form submitted, collect email and password from form if (isset($_POST['login'])) { $email = $_POST['email']; $password = $_POST['password']; // Check if a user exists with given username & password $result = mysqli_query($conn, "select 'Email', 'Password' from tblstudent where Email='$email' and Password='$password'"); // Count the number of user/rows returned by query $user_matched = mysqli_num_rows($result); // Check If user matched/exist, store user email in session and redirect to sample page-1 if ($user_matched > 0) { $_SESSION["email"] = $email; header("location: welcome.php"); } else { echo "User email or password is not matched <br/><br/>"; } } ?> Edited May 4 by Barand code tags added i have already made the register page where their info goes into the database, and im not sure about the code that selects values from the database. mysql_connect('', '', ''); mysql_select_db(''); $user = $_POST['user']; $pass = $_POST['pass']; echo "<font color='white'>You Need To Login</font>"; if($user == Username && $pass == Password) echo "Welcome $user"; mysql_query("SELECT ('Username', 'Password') FROM login"); ?> Hello again guys, am having trouble with a login script I have been working on. Have crawled the web for answers so thought i would post here again for some help. A brief run down on what the script is intended to do: 1.) The script checks to see if a user is logged in, and asks them to if their not. 2.) If theuser is logged in their userid is grabbed from $_SESSION and assigned to $userid 3.) Connection to the database is made and the field premium is updated with value "1" where userid = $userid A error message is supposed to be shown if the query has an error, but currently an error is not produced, the premium field remains NULL and the echo is shown. Can't for the the life of me fiqure out why it isn't working, but i think it is quite simple. Heres my script <?php if (!is_authed()) { echo 'You are not logged-in. Please login so we can add your purchased video to your account.'; include 'login_form.inc.php'; } else { include 'cp/config.php'; include 'cp/opendb.php'; $_SESSION['userid'] = $userid; $query = "UPDATE user SET premium='1' WHERE userid='$userid'"; mysql_query($query) or die('Error, query failed : ' . mysql_error()); echo 'Thank You for purchasing our series. We have added it to your account so you can use it straight away.'; } ?> For some reason the Obviously when connecting to php Im not going to show all of my login details; mysql_connect("details","details","password") or die(mysql_error()); mysql_select_db("details") or die(mysql_error()); whats the best way to hide them? Ive seen some people using an include file with their login details on but say for eg. <?php include('con.php'); ?> Whats to stop somone looking at www.myweb/con.php and obtaining my details there instead? Ok so I need to create a form to accept the users EmailAddress and Password as credentials to your site then use an SQL Query to determine if the person has an account Code: [Select] <?php require "connectionInfo.php"; $error = ""; if(!isset($_POST["personId"]) || !isset($_POST["firstName"]) || !isset($_POST["lastName"]) || !isset($_POST["emailAddress"]) || !isset($_POST["telephoneNumber"]) || !isset($_POST["socialInsuranceNumber"]) || !isset($_POST["password"]) ) { $error = "Please fill in the info"; } else { if($_POST["personId"] != "" && $_POST["firstName"] != "" && $_POST["lastName"] != "" && $_POST["emailAddress"] != "" && $_POST["telephoneNumber"] != "" &&$_POST["socialInsuranceNumber"] != "" && $_POST["password"] != "") { $dbConnection = mysql_connect($host, $username, $password); if(!$dbConnection) die("Could not connect to the database. Remember this will only run on the Playdoh server."); mysql_select_db($database); $sqlQuery = "INSERT INTO persons (personId, FirstName, LastName, emailAddress, telephoneNumber, socialInsuranceNumber, password) VALUES('".$_POST["personId"]."', '".$_POST["firstName"]."', '".$_POST["lastName"]."', '".$_POST["emailAddress"]."', '".$_POST["telephoneNumber"]."', '".$_POST["socialInsuranceNumber"]."', '".$_POST["password"]."')"; if(mysql_query($sqlQuery)) $error = "Person Successfully Added"; else $error = "Person Could not be added ".mysql_error(); mysql_close($dbConnection); } else $error = "Please enter all the information"; } ?> <form action="createAccount.php" method="post"> Person ID: <input type="text" name="personId" /> <br /> First Name: <input type="text" name="firstName" /> <br /> Last Name: <input type="text" name="lastName" /> <br /> Email: <input type="text" name="emailAddress" /> <br /> Telephone: <input type="text" name="telephoneNumber" /> <br /> Social Insurance Number: <input type="text" name="socialInsuranceNumber" /> <br /> Password: <input type="text" name="password" /> <br /> <input type="submit" value="Submit to Database" /> </form> -----EDIT----- Ok I was able to create the html code for it, but how do I use an sql query to determine if the person has an account? Code: [Select] <form method='post' action='login.php'> <table><tr><td>Email Address:</td><td><input type='text' name='emailAddress'></td></tr> <tr><td>Password:</td><td><input type='password' name='password'></td></tr> <tr><td></td><td><input type='submit' name='submit' value='Log in'></td></tr></table> </form> This topic has been moved to PHP Installation & Configuration. http://www.phpfreaks.com/forums/index.php?topic=355401.0 Hi All, First time posting here. I've googled the problem, but can't seem to find a response that's the same. All I want to do is have a list of id numbers and for each id number in the array, submit a MySQL query to retrieve information relating to the id number. When I execute the code below however, I end up with only the last item in the array being printed in the echo statement. Any clues? Thanks, Code: [Select] // get array of ids $ids = getIDs($ids); // loop through input list foreach ($ids as &$id) { getVarDetails($id); } function getVarDetails($local) { $con = mysql_connect('localhost:3306', 'root', '********'); if (!$con) { die('Could not connect: ' . mysql_error()); } // set database as Ensembl mysql_select_db("Ensembl", $con); $result = mysql_query("SELECT * FROM variations WHERE name = '$local' LIMIT 1"); $row = mysql_fetch_array($result) while($row = mysql_fetch_array($result)) { echo $row['name'] . " " . $row['id']; echo "<br />"; } // close connection mysql_close($con); } I am very new to php and am trying to create a simple application that uploads a PDF file to a database. I have one field for the Volume Number and on file field for the PDF to be uploaded. My issue is i can't get the PDF to upload or insert the name of the pdf (eg volume1.pdf) into the data base. I would also like to point out that I know i have a low post count, but i only seek help when i truly need it and have exhausted all other resources... Here is what i have, please go easy on me this is my first round at php: Code: [Select] <?php if(isset($_POST['submit'])){ $vol_num = $_POST['vol_num']; $pdf = $_FILES['pdf']['name']; $path = '../pdf/'.$_FILES['pdf']['name']; move_uploaded_file($_FILES["pdf"]["tmp_name"], $path); mysql_query("INSERT INTO volumes set vol_num='$vol_num', vol_link='$pdf'") or die (mysql_error()); echo "<script>location.href='add_volume.php'</script>"; } ?> what am i doing wrong here? Thanks in advance |