I'm trying to create a system on my HTML website where you can enter a code which then gets checked by a php-file. The php shall compare the code with codes that are in my MYSQL Table. The php shall also check whether the code was allready used or not. After the check is completed a message shall appear on the Website that tells the user if his code is valid or not.
If the Code is valid, it shall inform me somehow that somebody entered a code.
I've wrote some code but when I press the button it's not working at all.Here my HTML code:
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11/jquery.min.js">
</script>
<script>
$(document).ready(function() {
//the min chars for promo-code
var min_chars = 6;
//result texts
var checking_html = 'Checking...';
//when keyup
$('#code').keyup(function(event){
//run the character number check
if($('#code').val().length == min_chars){
//show the checking_text and run the function to check
$('#Promo_code_status').html(checking_html);
check_code();
}
});
});
//function to check the promo code
function check_code(){
//get code
var code = $('#code').val();
//use ajax to run the check
$.post("check_code.php", { code: code },
function(result){
//if the result is 0
if(result == 0){
//show that the code is correct
$('#Promo_code_status').html(code + ' is correct.');
}else if(result == 1){
//show that the code is correct, but already has been used
$('#Promo_code_status').html(code + ' has allready been used.');
}else{
//show that the code is not correct
$('#Promo_code_status').html(code + ' is not correct.');
}
});
}
</script>
</head>
<body>
<center><h1>Enter Promo Code</h1>
<form method="post" action="check_code.php">
<input type="text" id="code" name="code" maxlength="6" />
<div id="promo_code_status"></div>
<br>
<input type="submit" value="Let's go"></center>
</form>
</body>
</html>
And here you have my PHP File:
<?php
//connect to database
$user = "***"; //Username
$pass = "***"; //Password
$host = "localhost"; //Host
$dbdb = "***"; //Database name
$connect = mysqli_connect($host, $user, $pass, $dbdb);
if(!$connect)
{
trigger_error('Error connection to database: '.mysqli_connect_error());
}
//get the code
mysqli_real_escape_string($connect, $_POST['code']);
//mysql query to select field code if it's equal to the code that we checked '
$result = mysqli_query($connect, 'select Code from Codes where Code = "'. $code .'"');
$record = mysqli_fetch_array($result);
//if number of rows fields is bigger them 0 that means the code in the database'
if(mysqli_num_rows($result) > 0){
if($record['used'] == 0) {
//and we send 0 to the ajax request
echo 0;
} else{
//and we send 1 to the ajax request
echo 1;
}
}else{
//else if it's not bigger then 0, then the code is not in the DB'
//and we send 2 to the ajax request
echo 2;
}
?>
By pressing the Button "Let's go" the PHP shall do it's Job and afterwards it shall give the user the information about his code.
What do I have to change?
I thing you not assign the post value in $code, that's reason.
$code = mysqli_real_escape_string($connect, $_POST['code']);
make sure the jquery CDN that you are using can be accessed. because it looks like wrong CDN.
in the checkcode() function, you use #Promo_code_status, P is uppercase, while in html you use promo_code_status with p is lowercase.
in php file you forget to assign values to the $code
I hope this helps
Related
I want to validate my form using JQuery and use php to send it to my database.
This is what I've tried:
<body>
<form id="first_form" method="post" action="">
<div>
<label for="first_name">First Name:</label>
<input type="text" id="first_name" name="first_name"></input>
</div>
<div>
<label for="last_name">Last Name:</label>
<input type="text" id="last_name" name="last_name"></input>
</div>
<div>
<label for="handphone">Handphone:</label>
<input type="text" id="handphone" name="handphone"></input>
</div>
<div>
<input type="submit" value="Submit" />
</div>
</form>
<script>
$(document).ready(function() {
$('#first_form').submit(function(e) {
e.preventDefault();
var first_name = $('#first_name').val();
var last_name = $('#last_name').val();
var handphone = $('#handphone').val();
$(".error").remove();
if (first_name.length < 1) {
$('#first_name').after('<span class="error">This field is required</span>');
return false;
}
if (last_name.length < 1) {
$('#last_name').after('<span class="error">This field is required</span>');
return false;
}
if (handphone.length < 1) {
$('#handphone').after('<span class="error">This field is required</span>');
return false;
}
});
});
</script>
<?php
$first_name = "<script>var first_name = $('#first_name').val();</script>";
$last_name = "<script>var first_name = $('#last_name').val();</script>";
$handphone = "<script>var first_name = $('#handphone').val();</script>";
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO vali (first_name, last_name, handphone)
VALUES (first_name, last_name, handphone)";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
</body>
So as you can see, the first part is just the html and form.
The second part is where I used Jquery to validate (which works).
Now the issue is at the php part, where it sends the data to my database as empty.
The logic behind my php code is that, I take the values of the variables I set in Jquery and just update it to the database. But its updating empty values.
I tried to return false in Jquery, but its not working for me. May I know what am I doing wrong here?
Firsty, you have to understand JavaScript in this case is a client-scripting language and PHP is a server-scripting language. Variables can't be pass from JavaScript to PHP is the <script> tag.
Secondly, we have different types of requests; GET, POST, PUT, DELETE and so on. This we can check via PHP. A form can only send two types of request as at the time of me typing this answer, GET and POST request, this is the method value you set in your form. Every value set in a form can be accessed in the request global array and the name of the value being the key/indetifier. $_GET[] if you sent a get request and $_POST[] if you sent a post request.
In your form above, your method is set to POST, meaning all the values would be available in the global $_POST[].
When ever a page loads, the server script gets loaded first before the Client Side / HTML / CSS and other resources used by your document.
PHP has a function called isset() which is used to check if a variable available regardless it's value which is very useful for form validation.
If you ever have PHP codes in file performing logic/operations, it is advisable you placed them topmost of you file since it isn't used in rendering your document/view.
From your code, you should edit as
<?php
if(isset($_POST['submit'])){
// This block will only be executed when a submit button is triggered
//Make your connection here, it is not a must but good practice to always make connection first in order to make it available everywhere. Ensure to check if connection was successful or not.
$first_name = $_POST['first_name']; // this gives you the exact value you passed in the form.
...
// All you variables should be assigned using the above method
//in the above assignation of variables, we didn't sanitize. mysqli has a method real_escape_string() which does that for us. Ensure to always sanitize and properly validate inputs you send to your DB
//Sanitizing inputs
$first_name = $con ->real_escape_string($first_name);
// Do the above for other inputs, then you are good to perform an insert.
$result = $conn->query($sql);
if($result){
//Record has been inserted, you can choose to redirect or do some other logic
}else{
//Record was not successfully inserted. Print your error if in development to get an insight to what went wrong
}
}
$conn->close()
?>
I'm having trouble inserting my query into the database after the program checks the username availability on blur using ajax to check if the username is available for register or not. After it checks the username availability, it always insert some empty queries in my database.
Here is the sample output for more details:
But the problem is when it displays username available it will automatically inserts an empty query into the database.
This will insert into the database after it checks the availability of the user:
I want the insert query to be inserted when the user submits or when he/she clicks the register button. The availability of the username is for checking only. Is there a problem with my code?
Here is the code:
Sample.php FILE
<!DOCTYPE html>
<html>
<head>
<title>Sample</title>
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>
<input type="text" placeholder="Username" id="username" name="username" class="form-control" >
<span id="availability" style="color:green;"></span>
<input class="form-control" placeholder="First Name" name="firstname" type="text" >
<input type="submit" id="signup" name="signupsubmit" class="btn btn-info form-control" value="Register">
</body>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script>
//Check Username availability
$(document).ready(function()
{
$('#username').blur(function()
{
var username = $(this).val();
$.ajax({
url:"SampleProcess.php",
method:"POST",
data:{username:username},
dataType:"text",
success:function(html)
{
$('#availability').html(html);
}
})
});
});
</script>
</html>
I use ajax for checking the username availability inside the script here:
<script>
//Check Username availability
$(document).ready(function()
{
$('#username').blur(function()
{
var username = $(this).val();
$.ajax({
url:"SampleProcess.php",
method:"POST",
data:{username:username},
dataType:"text",
success:function(html)
{
$('#availability').html(html);
}
})
});
});
</script>
The process file for checking username and inserting a query.
SampleProcess.php FILE
<?php
//This is an external file DBController for creating a connection to DB
require_once("DBController.php");
//DBHandler handles the DBController class
$DBHandler = new DBController();
//Call the function mainConnect()
$connect = $DBHandler->mainConnect();
//Check Username Availability
if(isset($_POST["username"]))
{
$sql = "SELECT * FROM tablereminders WHERE username ='".$_POST["username"]."'";
$result = mysqli_query($connect, $sql);
if(mysqli_num_rows($result) > 0 )
{
echo"<span class=\"text-danger\">Username not available</span>";
echo'<script>document.getElementById("username").focus();</script>';
}
else
{
//In this else condition the query will be inserted in the
//database after the user clicks the register button, but it will insert right
//after the program check the availability
echo"<span class=\"text-success\">Username available</span>";
//Problem inserting my query here
$Username = isset($_POST['username']);
$FirstName = isset($_POST['firstname']);
//Query Insert into tablereminders
$query = mysqli_query($connect, "INSERT INTO `tablereminders`(`username` , `firstname`)
VALUES ('$Username' , '$FirstName')");
if($connect->query($query) == TRUE)
{
echo "<script type='text/javascript'>alert('New record created successfully');</script> ";
echo "<script>setTimeout(\"location.href = 'LoginReminder.php'\", 0)</script>";
}
}
}
?>
In the else condition after checking mysqli_num_rows I put my insert query there to perform when the user clicks the register buton. Is there a problem with my if-else condition or is it about the AJAX function inside the Sample.php file, Should I use GET or POST method? Either way I tried it but it doesn't correct my problem here. Please help.
Try this ;)
There are two things
Check username availability.
Submit the form and create a record.
1. Check username availability.
For this, you have done all the code you need to modify code like this:
<?php
//This is an external file DBController for creating a connection to DB
require_once("DBController.php");
//DBHandler handles the DBController class
$DBHandler = new DBController();
//Call the function mainConnect()
$connect = $DBHandler->mainConnect();
//Check Username Availability
if(isset($_POST["username"])){
$sql = "SELECT * FROM tablereminders WHERE username ='" . $_POST["username"] . "'";
$result = mysqli_query($connect, $sql);
if(mysqli_num_rows($result) > 0){
echo '<span class="text-danger">Username not available</span>';
echo '<script>document.getElementById("username").focus();</script>';
}else{
echo '<span class="text-success">Username available</span>';
}
}
2. Submit the form and create a record.
Now create new ajax call to some other file to insert record which submits form data on some button click event.
This you can do same you done in case of username availability.
Code sample to insert record:
$Username = isset($_POST['username']);
$FirstName = isset($_POST['firstname']);
//Query Insert into tablereminders
$query = mysqli_query($connect, "INSERT INTO `tablereminders`(`username` , `firstname`)
VALUES ('$Username' , '$FirstName')");
if($connect->query($query) == TRUE){
echo '<script type="text/javascript">alert("New record created successfully");</script>';
echo '<script>setTimeout(function(){ location.href = "LoginReminder.php"}, 0);</script>';
}
Remember 2 things before inserting new record in database:
Validate user data.
Escape user data may be with this function mysqli_real_escape_string OR you can use prepared statement and bind params.
I am new to PHP and can't find answers as to why the following code doesn't work. This should be easy, but I can't figure this out. This code produces no errors, and the SQL statement is correct in the phpAdmin SQL console. I've searched web & StackOverflow, but can't find a good answer. What's wrong?
ALL users (whether in the db or not) get ignored and stuck on login page.
<?php
session_start();
//create function to check login form for admin or other type of user.
//Redirect the admin user to the welcome page.
function login()
{
//strip login and password using in-build htmlspecialchars function
$value1 = htmlspecialchars($_POST['login']);
$value2 = htmlspecialchars($_POST['password']);
//set variables for the db connection
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb";
$loggedin = '';
//Create new connection to db
$conn = new mysqli($servername, $username, $password, $dbname);
//Check connection and handle any error
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
header('Locatin: login.php');
}
else {
//check if super admin user exists in db
$sql = "SELECT count(*) FROM admins WHERE AdminLevel = 1";
$result = mysqli_query($conn,$sql);
//check to see if query returns any rows
if(mysql_num_rows(($result) > 0) {
include 'welcome.php';
}
//check if the password and username match
if(($username === $value1) && ($password === $value2)) {
$_SESSION['loggedin'] = TRUE;
echo "Hello ".$value1.", you are logged in!<br>";
}
//send user error message if login/username and password wrong
else {
echo "Incorrect username or password<br>";
include 'login.php';
}
//close the db connection
$conn->close();
}
?>
Login Form:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Admin Login</title>
<script>
//function to check the form
function chkForm()
{
//determine the number of elements in the user login form
var intFormLen = document.forms[0].elements.length;
//loop through the form fields to see that a value has been input
for (var i = 0; i < intFormLen; i++) {
if (document.forms[0].elements[i].value == "") {
//send user an error message if login field empty
document.getElementById(document.forms[0].elements[i].name).innerHTML="Required Field";
document.forms[0].elements[i].focus();
return false;
}
}
//clear the form fields
function clearWarn(fieldName)
{
document.getElementById(fieldName).innerHTML = "";
return true;
}
return;
}
</script>
</head>
<body>
<h2>Admin Login</h2>
<div class="phpEcho">
<div class="formLayout">
<form action="#" method="post" onsubmit="return chkForm();">
<label for="login">Login:</label>
<input type="text"name="login" onchange="return clearWarn('fieldName')">
<div id="login" style="color:red"></div><br>
<label for="password">Password:</label>
<input type="password" name="password" onchange="return clearWarn('fieldName')">
<div id="password" style="color:red"></div><br><br>
<input type="submit" name="cmdSubmit" value="Log in">
</form>
</div>
</div>
</body>
</html>
You set your form action="#" and don't submit it in JavaScript.
As noted by Jason, chkForm() will never return true, which would also prevent the form from submitting.
This script has a lot of issues that should be addressed. I will go over a couple things that may help you:
1) I would suggest using some kind of config / bootstrap file to include in your documents that contains reusable elements and start session. Require/include only once.
/config.php
define("_DS_",DIRECTORY_SEPARATOR);
define("DBUSER",'root');
define("DBPASS",'');
define("DBHOST",'localhost');
define("DBDATABASE",'mydb');
// Start session
session_start();
2) You will want to separate out your functions, importantly your database connection, whether by class or by function. You want to keep tasks separate so it's easy to reuse.
Here is an example (I am going to use PDO because I am more familiar with it but principle is the same):
/functions/connection.php
function connection()
{
// This is just a really basic connection, one could expand on this
return new PDO('mysql:host='.DBHOST.';dbname='.DBDATABASE, DBUSER, DBPASS);
}
/functions/login.php
/*
** #param $username [string] by making this a param, you can manually log in users outside of POST
** #param $password [string] same as username
** #param $conn [resource] You will want to inject your connection into this
** in order to use it. Don't make the connection
** inside. May as well reuse resources already active
** #return [bool] If you return TRUE or FALSE, that will tell your script
** whether the login succeeded or failed for notification
*/
function login($username,$password,$conn)
{
// Don't worry about stripping down the username/pass, just bind
// the username and match the password
// You need to select from your user table (or whatever table
// you are storing your usernames for your site)
$query = $conn->prepare("select * from `users` where `username` = :0");
$query->execute(array(':0'=>$username));
$result = $query->fetch(PDO::FETCH_ASSOC);
if(empty($result))
return false;
// You will want to use password_hash to save passwords
if(!password_verify($password,$result['password']))
return false;
// I use htmlspecialchars here so I don't forget when echoing to page
// but you can do it at the time you echo to browser
$_SESSION['first_name'] = htmlspecialchars($result['first_name']);
//etc....
return true;
}
To use:
/index.php
// Include our soon-to-be-used files
require_once(__DIR__._DS_.'config.php');
require_once(__DIR__._DS_.'functions'. _DS_.'connection.php');
require_once(__DIR__._DS_.'functions'. _DS_.'login.php');
// Set connection
$con = connection();
// See if a post has been made
if(isset($_POST['login'])) {
$loggedin = login($_POST['login'],$_POST['password'],$con);
}
// If the login attempt made
if(isset($loggedin)) {
// If successful
if($loggedin) {
header('Location: welcome.php');
exit;
}
else
// If failed, you can note in a variable an echo in the html section
$error = 'Login failed';
}
For the client-side validation, I would suggest jQuery Validate, it's easy and works very well.
I have three files working on an login app to learn PHP.
This is the connection with DB
<?php
# Connecting database below
$connection = mysqli_connect('localhost','root','','loginapp');
if ($connection) {
# code...
echo "connected";
}
else{
echo "Errorr";
die("Database");
}?>
and here is the html code for the web view
<html>
<head>
<title>Form</title>
</head>
<body>
<h1>Welcome to My Form</h1>
<form class="" action="login_create.php" method="post">
<input type="text" name="name" placeholder="Enter your name here"><br>
<input type="password" name="password" placeholder="Enter Password" value=""><br>
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>
and here is the file where things are going wrong, its not checking the conditions of entries and not putting the data into database what's wrong going there? help please
sometimes it gives
error that "unknown 'sbumit' in the $_POST" and sometimes it don't
doesn't even show any error
but doesn't even do anything
<?php
include "db.php";
if (isset($_POST['submit'])) {
$username = $_POST['name'];
$password = $_POST['password'];
if (isset($username) && isset($password)) {
if (strlen($username) > 10 && strlen($username) < 3) {
echo "Must enter username & pass between 3 & 10";
echo "So that we can forward your request";
}
else {
$query = "INSERT INTO users (username,password) VALUES ('$username','$password')";
$result = mysqli_query($connection,$query);
if(!$result)
{
die('Sorry Query faild'.mysqli_error());
}
}
}
else
{
echo "You haven't wrote anything, write it first";
}
}?>
Habib,
Some guidance for PHP :
$button = isset($_POST["submit"])?$_POST["submit"]:"";
What this line does is apply a value to the $button variable, the first check is that IF isset($var) THEN (indicated with the ? ) apply the value of $var to the $button variable.
The colon : then sets that if the boolean query (true/false) of the IF returns false, then apply the second value instead, in this case an empty string of "".
This is code minimalisation and you should be aware of it but there is little need to use it, especially while learning.
Feedback on your code:
mysqli_error($connection); Your error feedback for MySQLi should include the connection details, as shown here.
replace the $username = $_POST['name'];
$password = $_POST['password'];
if (isset($username) && isset($password)) {
because you want to check not if they're set but if they're not empty, currently they will be set as they're set to the values of $_POST even if they are null (potentially), so replace with:
if(!empty($username) && !empty($password)){
Also note that ! is the negative operator. so above is IF NOT EMPTY.
if (strlen($username) > 10 && strlen($username) < 3) { this is impossible to reach because you're setting if string is longer then 10 AND string is shorter than 3, this is clearly impossible. replace the && with || which is OR rather than AND .
Personally I think that isset($_POST['submit']) is not the best way, instead checking that if($_POST['submit'] == 'submit') confirms the submission of this form from this submit button (the value is the value set in your HTML form).
$query = "INSERT INTO users (username,password) VALUES ('$username','$password')"; This works fine, BUT you really, really need to do some research into SQL injection attacks and SQL security. read How can I prevent SQL injection in PHP? as a start. This is very important to learn at the start of your PHP MySQL learning.
Also research into PDO database connectivity.
Also be aware that your script will not output anything when you have a successful saving of username/password to the database.
As a closer:
Fnally, set up error logging on your page, to give you useful feedback on errors and problems: error_reporting(E_ALL);
ini_set('display_errors', 1); at the very top of your page. Also see How do I get PHP errors to display?
Change your code as follow.
<?php
include "db.php";
$button = isset($_POST["submit"])?$_POST["submit"]:"";
$username = isset($_POST["name"])?$_POST["name"]:"";
$password = isset($_POST["password "])?$_POST["password "]:"";
/*Commetents*/
$button =isset($_POST["submit"])?$_POST["submit"]:"";
is similar to following code:
if(isset($_POST["submit"]))
{
$button = $_POST["submit"];
}
else
{
$button = $_POST["submit"];
}
You know in Php 5.4 , it will present error,if you do not set any value to variable . that is why we used it. If it doesn't get any value it will set it value "".
if($button == "submit") means when someone will press the button submit then $_POST['submit'] value will be submit which you define in the submit button value.
if($button == "submit")
{
if($username=="" or $password=="")
{
$error ="Username & Password can't be blank";
}
elseif(strlen($username)<3 or strlen($username) > 10 )
{
$error ="Must enter username & pass between 3 & 10";
}
else
{
$query = "INSERT INTO users (username,password) VALUES('$username','$password')";
mysqli_query($connection,$query) or die(mysqli_error());
}
}
echo $error;
Hope it will help you .
I have the following scripts. The user will pass a numerical value e.g. 123 as a parameter in the URL, and the app will load/fetch that value from MySQL and show it in the textarea.
E.g., entering, "example.com/index.php?id=123" in the URL will pull the 123rd record from the database and show it in the textarea. Everything seems to be working, but I want to show the last row/id of the "source" table when the form is submitted. So, instead of showing "Saved!" in the pop-up, it should show something like "124" (or, the last updated row/number).
index.php:
<?php
include('connect-db.php');
if (isset($_GET['id']) && is_numeric($_GET['id']) && $_GET['id'] > 0)
{
$id = $_GET['id'];
$result = mysql_query("SELECT content FROM source WHERE id=$id") or die(mysqli_error());
$row = mysql_fetch_array($result);
if($row)
{
$submit_date = date("Ymd");
$content = $row['content'];
}
}
?>
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#myform").submit( function () {
$.post(
'submit.php',
$(this).serialize(),
//
// show the last id here!
//
function(data){ alert("Saved!"); }
);
return false;
});
});
</script>
</head>
<body>
<div id="header" >
<form action="submit.php" method="post" id="myform">
<textarea id="editor" name="editor" id="editor"><?php echo $content; ?></textarea>
<br/>
<input type="submit" name="submit" value="Submit" id="submit"/>
</form>
</div>
</body>
</html>
submit.php:
<?php
include('connect-db.php');
$submit_date = date("Ymd");
$content = mysql_real_escape_string(htmlspecialchars($_POST['editor']));
$ip_address = $_SERVER['REMOTE_ADDR'];
if ($content != '') {
//
// Show the variable (last_id) value in the JavaScript above!
//
$last_id = mysql_insert_id();
mysql_query("INSERT INTO source (submit_date, ip, content) values('$submit_date','$ip_address','$content')") or die(mysql_error());
mysql_close($connection);
}
?>
connect-db:
<?php
$server = 'localhost';
$user = 'domdom';
$pass = 'password#123';
$db = 'domdom';
$connection = mysql_connect($server, $user, $pass) or die ("Could not connect to server ... \n" . mysql_error ());
mysql_select_db($db) or die ("Could not connect to database ... \n" . mysql_error ());
?>
Please note that I do not need any advice on PDO or MySQLi, that is not important at this time. However, any suggestions on improving the security/SQL query etc. are welcome. Thanks.
I will try to answer in 2 parts:
[1] PHP side: in the "isset($_GET['id'])" section, just return the content of the text area you want to fill out. Say, in your DB, 123 -> 'Friday'. Just return the String 'Friday'.
[2] Client side: Do not "submit" the form. Call a JS function, and in there, create an XmlHttpRequest, and send the request to the server as an AJAX request. When you get the return value, examine it and populate your HTML textarea.
If you think this method is suitable for you and you need more details, let me know. I do this in many of my sites.