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PHP - Users Are Still Registered Into My Database Even Though They Enter A Password
less than 6 characters. I think it's the way my code is ordered. I've tried switching the commands around, no luck.
Help please. Code: [Select] <?php //begin register script $submit = $_POST['submit']; //form data $username= strip_tags ($_POST['username']); $email= strip_tags($_POST['email']); $pwd= strip_tags($_POST['pwd']); $confirmpwd= strip_tags($_POST['confirmpwd']); $date = date("Y-m-d"); if ($submit) { //check for required form data if($username&&$pwd&&$confirmpwd&&$email) { //check length of username if (strlen($username)>25||strlen($username)<6) { echo "<p class='warning'>username must be bewteen 6 and 25 characters</p>"; } else { //check password length if (strlen($pwd)>25||strlen($pwd)<6) { echo "<p class='warning'>password must be between 6 and 25 characters</p>"; } else { //register the user echo "<p class='success'>Thanks for signing up!</p>"; } } //check if passwords match if ($pwd==$confirmpwd) { } else { echo "<p class='warning'>your passwords do not match</p>"; } //encrypt password $pwd = md5($pwd); $confirmpwd = md5($confirmpwd); //open database $connect = mysql_connect("xxxxxxxx", "xxxxxxxx", "xxxxxxxx"); mysql_select_db("digital"); //select database //register the user $queryreg = mysql_query(" INSERT INTO users VALUES ('','$username', '$email', '$pwd') "); die("<p class='success'>Thank you for signing up you have been registered"); } else { echo "<p class='warning'>please fill in all fields</p>"; } } ?> Similar TutorialsI'm using this type of code below. how can my site may have access to my e-mail ? and how can I send e-mails containing passwords to users.?? please help. //code is like this// if($_POST['submit']=='Register') { // If the Register form has been submitted $err = array(); if(strlen($_POST['username'])<4 || strlen($_POST['username'])>32) { $err[]='Your username must be between 3 and 32 characters!'; } if(preg_match('/[^a-z0-9\-\_\.]+/i',$_POST['username'])) { $err[]='Your username contains invalid characters!'; } if(!checkEmail($_POST['email'])) { $err[]='Your email is not valid!'; } if(!count($err)) { // If there are no errors $pass = substr(md5($_SERVER['REMOTE_ADDR'].microtime().rand(1,100000)),0,6); // Generate a random password $_POST['email'] = mysql_real_escape_string($_POST['email']); $_POST['username'] = mysql_real_escape_string($_POST['username']); // Escape the input data mysql_query(" INSERT INTO tz_members(usr,pass,email,regIP,dt) VALUES( '".$_POST['username']."', '".md5($pass)."', '".$_POST['email']."', '".$_SERVER['REMOTE_ADDR']."', NOW() )"); if(mysql_affected_rows($link)==1) { send_mail( 'myemail@xx.net', $_POST['email'], 'sitename - Your New Password', 'Your password is: '.$pass); $_SESSION['msg']['reg-success']='We sent you an email with your new password!'; } else $err[]='This username is already taken!'; } if(count($err)) { $_SESSION['msg']['reg-err'] = implode('<br />',$err); } header("Location: demo.php"); exit; } //but while running it on local server it shows the message "failure" as I have mentioned in the index.php. please provite a template codeing to solve the problem// Hello Everyone, I recent made a simple membership website. Every page I created works exactly how I envisioned it... All members data from my registration form goes into my database along with their md5 Encrypted passwords with a time-stamp. Subsequent pages have a start_session included. I am very please with it except ONE THING. Logging in is now a problem... username is recognized but NOT the password. Now the strange thing is that when I go into the database and copy the encrypted password and paste it into the password field in my login page, I miraculously get into my website with NO problem. " How do I get the registered members Encrypted Passwords to be recognized by the database when the registered members decide to logging in with the password that they create? " Is there a easy fix for this? I appreciate ALL your help... thanks mrjap1 Hi, I would like users to be able to send me payments though my website, but where they enter their own amount. I know this can be done using donations, but I do not want the word "donate" or "donations" to appear anywhere in the transaction. Can this be done at all? I have developed a code for a login and seems to work well (No syntax error according to https://phpcodechecker.com/ but when I enter a username and a password in the login form, I get an error HTTP 500. I think that everything is ok in the code but obviously there is something that I am not thinking about. The code (excluding db connection): $id="''"; $username = $_POST['username']; $password = md5($_POST['password']); $func = "SELECT contrasena FROM users WHERE username='$username'"; $realpassask = $conn->query($func); $realpassaskres = $realpassask->fetch_assoc(); $realpass= $realpassaskres[contrasena]; $func2 = "SELECT bloqueado FROM users WHERE username='$username'"; $blockedask = $conn->query($func2); $blockedres = $blockedask->fetch_assoc(); $bloqueado = $blockedres[bloqueado];
//Login if(!empty($username)) { // Check the email with database Hi! I have read like crazy to find a tutorial on a login page without My_SQL. Anyway I am working on a easy login/logged out page with sessions. Here is the login page with tree users in an array.
The things that I need some hints to solve is, when clicking on login the error message don't show. Instead the script goes to the logged in page right away. And when you write the wrong password you get loged in anyway.
I am not sure how or if it's possible to write a varible to a file this way. But I tried and recived a parse error with the txt varible.
When searching for topics I get more confused with the My_SQL varibles. I am near a breaking point at cracking the first step on PHP, but need some advice.
<?php
$page_title = 'Logged in'; //Dynamic title
include('C:/wamp/www/PHP/includes/header.html');
?>
<?php
session_start();
//A array for the sites users with passwords
$users = array(
'Dexter'=>'meow1',
'Garfield'=>'meow2',
'Miro'=>'meow3'
);
//A handle to save the varible users to file on a new line from the last entry
$handle = fopen("newusers.txt, \n\r")
$txt = $users;
fclose($handle);
if(isset($_GET['logout'])) {
$_SESSION['username'] = '';
header('Location: ' . $_SERVER['PHP_SELF']);
}
if(isset($_POST['username'])) {
if($users[$_POST['username']] == $_POST['password']) {
$_SESSION['username'] = $_POST['username'];
}else {
echo "Something went wrong, Please try again";
}
}
?>
<?php
echo "<h3>Login</h3>";
echo "<br />";
?>
<!--A legend form to login-->
<fieldset><legend>Fill in your username and password</legend>
<form name="login" action="777log.php" method="post">
Username: <br />
<input type="text" name="username" value="" /><br />
Password: <br />
<input type="password" name="password" value="" /><br />
<br />
<input type="submit" name="submit" value="Login" />
</fieldset>
</form>
<?php //Footer include file
include('C:/wamp/www/PHP/includes/footer.html');
?>
The logged in page
<?php //Header
$page_title = 'Reading a file';
include('C:/wamp/www/PHP/includes/header.html');
?>
<?php
session_start();
//Use an array forthe sites users
$users = array(
'Dexter'=>'meow1',
'Garfield'=>'meow2',
'Miro'=>'meow3'
);
//
if(isset($_GET['logout'])) {
$_SESSION['username'] = '';
echo "You are now loged out";
//The user is loged out and returned to the login page
header('Location: ' . $_SERVER['PHP_SELF']);
}
if(isset($_POST['username'])) {
//Something goes wrong here when login without any boxes filled
if($users[$_POST['username']] == $_POST['password']) {
$_SESSION['username'] = $_POST['username'];
}else {
echo "Something went wrong, Please try again";
$redirect = "Location: 777.php";
}
}
?>
<?php if($_SESSION['username']): ?>
<p><h2>Welcome <?=$_SESSION['username']?></h2></p>
<p align="right"><a href="777.php">Logga ut</a></p><?php endif; ?>
<p>Today Ben&Jerrys Chunky Monkey is my favorite!</p>
<?php //Footer
include('C:/wamp/www/PHP/includes/footer.html');
?>
I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
Can someone tell me why my php code isn't inserting users into the database. My database is made and table is made. I created a register form so my visitors can register but im getting a blank page. I has something to do with the following code
$statement = $connection->prepare('INSERT INTO users (username, email, password) VALUES (:usernamebox, :emailbox, :passwordbox)');
if ($result) {
Why is there the " : " before the usernamebox
Heres how my form looks like
<form action="register-clicked.php" method="POST"> Not sure how to even search for what I'm looking for, which is to say I'm really in need of guidance. I have a pre-populated database that I set up with about 260 names in it. It also has information like, height (basketball players) and high school. It's a list of kids I mean to invite to play in a league. From there, they will actually come to the league's site and register. I'd like for my form to check to see if they are already in the database, and if they are just add personal information, like email and phone number. If they are not, I'd like for the form to enter all of the information (name, school, etc). The company I work for is a not-for-profit organization, we provide medical services for people who can not afford to go to the regular doctor. We need a way to keep track of our patients and their visits and a way to search the data. I found an html form, with name, address, phone number, zip code,,,,,, and made a few changes to fit our needs. But I do not know how to get that information into the database. I also need a way to make sure the information is not a duplicate. Since we are a not-for-profit organization, funds are kinda tight. That is why I want to go with php instead of buying medical software. Most of the medical software on the market cost more then we can afford. I'am a computer tech here at the office, but my field is hardware, networking, and malware removal. PHP programming is new to me. So where do I start? How can I build a form with a drop down menu with the year, day, month? And then send that information to a database. If there is another/better option besides a drop down menu? If so, please tell me. The rest of the form is working fine, and all the inforamtion is going to the database - last name, first name, address, phone number, zip code, race,,,,,,,,,. I work for a not-for-profit medical office - we provide medical services to the local people who can not afford to go to a regular doctor. We need a simple way to keep track of patient information. One of the things we need to enter is the birthdate. Searching the forums provided a lot of results, but the words date and form are used so much, I can not find anything that helped me. I also looked through php.net, but can not find anything there. Sir/ma'am,
With the script I'm using to run my website, I've been trying to add an additional feature for the users to add/edit. I'll try to provide as much info as I can, hopefully it'll help.
Here is the code I'm using to display the user's unique info from the db.
<a class="wallet-edit"><?php echo $_SESSION['simple_auth']['INFO']?></a>That displays the user's info from the column 'INFO' perfectly. It's also a js popup to a menu to where I'm hoping to add a single textbox to edit the INFO. The script uses a similar function to edit the password with a popup. I've tried modifying the code to edit the INFO column but it doesn't work. Here is the default code it has to edit the password. I'm not sure if it can be changed to edit another column or needs a new piece of code for that.
// user edit
$('body').on('click', '.username-edit', function() {
$('#modal').html(' ');
var output = '<div class="modal-content"><h5><?php echo lang::get("Change password")?></h5><hr />';
output += '<h5><?php echo lang::get("New password:")?></h5><input type="password" name="password" id="password" value="" class="text ui-widget-content ui-corner-all" />';
output += '<h5><?php echo lang::get("Confirm password:")?></h5><input type="password" name="password2" id="password2" value="" class="text ui-widget-content ui-corner-all" />';
output += '</div>';
output += '<div class="modal-buttons right">';
output += '<button id="confirm-button" type="button" class="nice radius button"><?php echo lang::get("Change")?></button>';
output += '</div>';
output += '<a class="close-reveal-modal"></a>';
$('#modal').append(output);
$('#second_modal').hide();
$('#modal').reveal();
$('#confirm-button').click(function(){
$('#password').css('border-color', '#CCCCCC');
$('#password2').css('border-color', '#CCCCCC');
var password = $('#password').val();
var password2 = $('#password2').val();
if(typeof(password) === 'undefined' || password == ''){
$('#password').css('border-color', 'red');
return false;
}
if(password != password2){
$('#password2').css('border-color', 'red');
return false;
}
password_data = encodeURIComponent(password);
$.post("<?php echo gatorconf::get('base_url')?>", { changepassword: password_data} ).done(function(data) {
// flush
window.location.href = '<?php echo gatorconf::get('base_url')?>';
});
});
});
If the code above can be edited to work with what I'm trying to do, it of course only needs one textbox and doesn't have to be confirmed by a second input.
Please help! Thanks!
When a user logs in to my members area, their online status is set to 2 in my database and they are displayed as online for everyone else to see. And when they click log out their online status is set to 1 and they are no longer shown as online. My problem is when the session expires due to inactivity, the database isnt updated and they are still shown as online. So whenever the access a page in the members area I have made it so that time also gets entered into the database. And my goal is to have the users online page auto refresh every 5 minutes to check the current time against the time stored in the database for that user and if 15 minutes has past, have the online status in the database updated to 1 again. im using pdo but im a novice at it and havnt been able to find much help on google. Is there a way i can use an if statement to check the two times and only update the online field to 1 for the inactive users while leaving the users who appear active as 2? Sorry if this was confusing, i wasnt too sure how to word it properly. Hi there everyone. I'm really confused about this as I want to email eveyone in my DB when a new Event is created by a member. I have a test code which I used to make sure it works and it sends to 5 email addresses (which is all of them) in the test table. This works fine. All members with a 0 get the mail. So I now just took the code and changed the table name (obviously!!) and put it in my live site which has 50 members so far. It now doesn't send any emails??? Not sure why this is as the code is exactly the same (except the table name). $result = mysql_query("SELECT email FROM membership WHERE send_as_email = '0'"); while ($row = mysql_fetch_array($result)) { sendMail($row[0]); } mysql_free_result($result); function sendMail($to){ $subject = 'New Event Created; $message = "Hi there,\n Just letting you know of a new Event that has been created.\n Title: $title\n Town: $town\n Date: $date_convert\n For more details or to attend this Event, please login \n Many thanks\n The Team\n \n ****************************************************************************\n THIS EMAIL IS AUTOMATED. DO NOT REPLY.\n To turn notifications off, change the notification settings in your Profile.\n ****************************************************************************\n"; $headers = 'From: members@mysite.co.uk' . "\r\n" . 'X-Mailer: PHP/' . phpversion(); mail($to, $subject, $message, $headers); } Hey All, Been banging my head into the wall on this one. I have 2 tables one for users 'myMembers' and one for products 'products'. Each table has a auto increment id. The myMembers id is the user id and the products table id is for the product id. I have a row in the products table for agent_id. I would like the agent_id to be filled with the id from the myMembers table. I took a look at the manual; still do not understand how to take the id from the myMembers table then place that id into the agent_id; so the products(id) can be listed under the specific members id(agent_id) in the products table. So far my script for the products table inserts items correctly, but does not file under the specific agent_id. Here is the script for entering items to the products table. Thanks for the guidance! <?php // Script Error Reporting error_reporting(E_ALL); ini_set('display_errors', '1'); ?> <?php // Delete Item Question to Admin, and Delete Product if they choose if (isset($_GET['deleteid'])) { echo 'Do you really want to delete product with ID of ' . $_GET['deleteid'] . '? <a href="inventory_list.php?yesdelete=' . $_GET['deleteid'] . '">Yes</a> | <a href="inventory_list.php">No</a>'; exit(); } if (isset($_GET['yesdelete'])) { // remove item from system and delete its picture // delete from database $id_to_delete = $_GET['yesdelete']; $sql = mysql_query("DELETE FROM products WHERE id='$id_to_delete' LIMIT 1") or die (mysql_error()); // unlink the image from server // Remove The Pic ------------------------------------------- $pictodelete = ("../inventory_images/$id_to_delete.jpg"); if (file_exists($pictodelete)) { unlink($pictodelete); } header("location: inventory_list.php"); exit(); } ?> <?php // Parse the form data and add inventory item to the system if (isset($_POST['product_name'])) { $product_name = mysql_real_escape_string($_POST['product_name']); $price = mysql_real_escape_string($_POST['price']); $category = mysql_real_escape_string($_POST['category']); $subcategory = mysql_real_escape_string($_POST['subcategory']); $details = mysql_real_escape_string($_POST['details']); // See if that product name is an identical match to another product in the system $sql = mysql_query("SELECT id FROM products WHERE product_name='$product_name' LIMIT 1"); $productMatch = mysql_num_rows($sql); // count the output amount if ($productMatch > 0) { echo 'Sorry you tried to place a duplicate "Product Name" into the system, <a href="inventory_list.php">click here</a>'; exit(); } // Add this product into the database now $sql = mysql_query("INSERT INTO products (product_name, agent_id price, details, category, subcategory, date_added) VALUES('$product_name','$price','$details','$category','$subcategory',now())") or die (mysql_error()); $pid = mysql_insert_id(); // Place image in the folder $newname = "$pid.jpg"; move_uploaded_file( $_FILES['fileField']['tmp_name'], "../inventory_images/$newname"); header("location: inventory_list.php"); exit(); } ?> <?php // This block grabs the whole list for viewing $product_list = ""; $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $product_name = $row["product_name"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $product_list .= "Product ID: $id - <strong>$product_name</strong> - $$price - <em>Added $date_added</em> <a href='inventory_edit.php?pid=$id'>edit</a> • <a href='inventory_list.php?deleteid=$id'>delete</a><br />"; } } else { $product_list = "You have no products yet"; } ?> Hello I'm trying to set up a user area for my site where it displays the current logged in users ranking and other information in the future.
<?
ini_set('display_errors', 1);
require_once "header.php";
$sql = "SELECT * FROM users WHERE username = ?";
if($stmt = mysqli_prepare($link, $sql)){
mysqli_stmt_bind_param($stmt, 's', $_SESSION['username']);
if(mysqli_stmt_execute($stmt)){
$info = mysqli_fetch_array($stmt);
echo "Current rank:" . $info['rank'];
} else {
echo "Can't find user";
}
}
mysqli_stmt_close($stmt);
?>
That's the code I currently have but it gives me the error "but get an error message of mysqli_fetch_array() expects parameter 1 to be mysqli_result" Hello folks I am new to php and I have been trying to put together a database that a user can search and choose from the results. I have managed to make this script by copying code from google searches and trial and error. The script so far has been tested and works. The hard part is the code for choosing from the results, I have tried some things but I have been far from the mark, the thing is I can't get my head around the problem, if the first field is a number which is unique to each row, how can I pick that up in a php argument. I have tried making the first field an href link to send that number to a different table which would collect the results of the users choices, but I'm just not sure what to put in the code. Could someone throw me a lifeline here I've searched for hours on google to find any code that looks like it would work with no luck. // Get the search variable from URL $var = @$_GET['a'] ; $trimmed1 = trim($var); //trim whitespace from the stored variable $var = @$_GET['b'] ; $trimmed2 = trim($var); $var = @$_GET['c'] ; $trimmed3 = trim($var); $var = @$_GET['d'] ; $trimmed4 = trim($var); $var = @$_GET['e'] ; $trimmed5 = trim($var); $var = @$_GET['f'] ; $trimmed6 = trim($var); //connect to your database mysql_connect("localhost","root",""); //(host, username, password) //specify database mysql_select_db("a2149809_MV") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "SELECT * FROM `table` WHERE `field1` LIKE \"%$trimmed1%\" AND `field2` LIKE \"%$trimmed2%\" AND `field3` LIKE \"%$trimmed3%\" AND `field4` LIKE \"%$trimmed4%\" AND `field5` LIKE \"%$trimmed5%\" AND `field6` LIKE \"%$trimmed6%\" order by `field1`"; $result=mysql_query($query); $num=mysql_num_rows($result); mysql_close(); <table width="100%" border=2 cellspacing=2 cellpadding=2> <tr><form name="form" action="" method="get"> <td colspan="6"><input type="submit" name="Submit" value="Search" /> </td> </tr> <tr> <td><input type="text" name="a" value="" size="4" /></td> <td><input type="text" name="b" value="" size="40" /></td> <td><input type="text" name="c" value="" size="3" /></td> <td><input type="text" name="d" value="" size="10" /></td> <td><input type="text" name="e" value="" size="10" /></td> <td><input type="text" name="f" value="" size="10" /></td> </form></tr> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"Field1"); $f2=mysql_result($result,$i,"Field2"); $f3=mysql_result($result,$i,"Field3"); $f4=mysql_result($result,$i,"Field4"); $f5=mysql_result($result,$i,"Field5"); $f6=mysql_result($result,$i,"Field6"); ?> <tr> <td><?php echo $f1; ?></td> <td><?php echo $f2; ?></td> <td><?php echo $f3; ?></td> <td><?php echo $f4; ?></td> <td><?php echo $f5; ?></td> <td><?php echo $f6; ?></td> </tr> <?php $i++; } ?> </table> When I set up MS SQL on my machine, for whatever reason I did not set up a UID or password. I can not get the php to connect with SQL because I do not have a UID and password on the db (which I know is not smart). I don't understand code enough to know how to manipulate this string below to remove the requirement. Hello,
Here is my current code:
index.php
<html>
<body>
<form id="hey" name="hey" method="post" onsubmit="return false">
Name:<input type="text" name="name">
<input type="submit" name="submit" value="click">
</form>
<table class="table table-bordered" id="update">
<thead>
<th>Name</th>
</thead>
<tbody>
<script type="text/javascript">
$("#hey").submit(function() {
$.ajax({
type: 'GET',
url: 'response.php',
data: { username: $('#name').val()},
success: function (data) {
$("#update").prepend(data);
},
error: function (xhr, ajaxOptions, thrownError) {
alert(thrownError);
}
});
});
</script>
</tbody>
</table>
response.php
<?php $username = $_GET['username']; echo "<tr>"; echo "<td>$username</td>"; echo "</tr>"; ?>This code works fine, it prints out the rows as what the user enters. Now what I want to do is, log all these entries to a mySQL database and also, display these rows over the site. i.e., any user who is online, should be seeing this without having to refresh the page too.. Real time updates in a way. How can I achieve that? Thanks! Hi there everyone, I have the below code (needs to be sanitized for SQL injection - i know ). Submitting to a MYSQL database. I wish to check on form submit if the users email address already exists, and if so display a simple error message (even just a windows error message) stating "the email address you have entered already exists". I don't really know where to start with this, or what the code should look like, so any help and direction would be massively appreciated. All i do know, is the email column is set to unique, and when i attempt to submit using an email i know exists the code appears to run successfully without spitting out any errors (i.e. the web url changes to the below php code) but the table doesn't update (which is correct). I just don't know how to then return the user to the form (preferably with all their info still entered) when this happens along with a nice error message... Kind regards, Tom. Code: [Select] <? $localhost="00.000.000.00"; $username="###"; $password="###"; $database="mfirst"; $firstname=$_POST['firstname']; $surname=$_POST['surname']; $dob="{$_POST['dobyear']}-{$_POST['dobmonth']}-{$_POST['dobday']}"; if (isset ($_POST['permissionnewsletter']) || (!empty ($_POST['permissionnewsletter']))) { $permissionnewsletter = "Yes"; } else { $permissionnewsletter = "No"; } $email=$_POST['email']; $userpassword=$_POST['userpassword']; $telephone=$_POST['telephone']; mysql_connect($localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $insertintocustomerdetail = "INSERT INTO customerdetail VALUES (NULL,'$firstname','$surname','$dob','$permissionnewsletter','$email','$userpassword','$telephone',now())"; mysql_query($insertintocustomerdetail); $addressline1=$_POST['addressline1']; $addressline2=$_POST['addressline2']; $cityortown=$_POST['cityortown']; $county=$_POST['county']; $postcode=$_POST['postcode']=strtoupper(@$_REQUEST['postcode']); $insertintoaddresstable = "INSERT INTO addresstable VALUES (NULL,LAST_INSERT_ID(),'$addressline1','$addressline2','$cityortown','$county','$postcode',now())"; mysql_query($insertintoaddresstable); mysql_close(); ?> Hi, I have two html made text boxes one that is called "name" and another that is called "regnum". I also have a submit button. They are both used to add data to a database. The name text box should add a name to the database under the "name" heading and the regnum should add a number under the "regnum" heading Here is the code for them: HTML Code: <form action="" method="post"> <p> Name: <input type="text" name="name"/> </p> <p> Regnum: <input type="text" name="regnum"/> </p> <p> <input type="submit" value="Add To Database" name = "submit2" /> </p> </form> Here is the PHP code that i am using for adding the user to the database: PHP Code: $host="localhost"; $username="root"; $password=""; $database="lab2"; mysql_connect("$host", "$username", "$password") or die(mysql_error()); mysql_select_db("$database") or die(mysql_error()); $name = $_POST['name']; $regnum = $_POST['regnum']; if(!$_POST['submit2']){ echo "Enter A Vaue"; }else{ mysql_query("INSERT INTO lab2('name', 'regnum') VALUES(NULL, '$name', '$regnum')") or die(mysql_error()); echo "User Added To Database"; } The problem i get with this is "Undefined Index name and regnum". I watched a video on youtube and this is how the guy did it but it worked for him and for some reason it doesn't work for me. Can anyone help?? Thanks. |
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