I am setting up a new website and want to be able to gather user input on the checkout page and echo the user input on the confirmation page. How do I perform this operation?
This is using html and php. In the past, I've tried using POST method, isset, $_Request functions.
This is the code for Confirmationpage.html.
<!DOCTYPE html>
<html>
<head>
<title>Confirmation page</title>
</head>
<body>
<h1>Confirmation page</h1>
<h2>Order Confirmed!</h2>
<?php
// define variables and set to empty values
$fName = $lName = $email = $phoneNumber = "";
if ( isset( $_POST['submit'] ) ) {
$fName = $_POST["fName"];
$lName = $_POST["lName"];
$email = $_POST["email"];
$phoneNumber = $_POST["phoneNumber"];
echo $fName . " " . $lName;
echo $email;
}
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
</body>
</html>
This is the code for Checkoutpage.html.
<!DOCTYPE html>
<html>
<head>
<title>Checkout page</title>
</head>
<body>
<h1>Checkout page</h1>
<form action="Confirmationpage.html" method="POST">
<input type="text" name="fName" placeholder="First Name"><br>
<input type="text" name="lName" placeholder="Last Name"><br>
<input type="email" name="email" placeholder="Email"><br>
<input type="number" name="phoneNumber" placeholder="Phone Number">
<br>
<input type="submit" value="Confirm" name="submit">
</form>
</body>
</html>
I expected the first name, last name and email to be echoed in the confirmation page however the actual result is not outputting the first name, last name and email.
try this,
echo $fName . " " . $lName;
echo '<br>';
echo $email;die;
Keep this in mind: your file should be .php
Related
I tried to create a form which will send an email and add data to MySQL data base. For email part because I work on localhost I used formsubmit. All good but there is a conflict. When I press on send only the email is send without any data added to my database. If I delete the action attribute from form data will be added but in this case I can't send any emails. Here is my file:
<?php
require 'config.php';
if(!empty($_SESSION["id"]))
{
$id= $_SESSION["id"];
$result = mysqli_query($conn,"SELECT * FROM tb_user WHERE id= $id");
$row = mysqli_fetch_assoc($result);
}
else
{
header("Location: login.php");
}
if(isset($_POST["submit"]))
{
$name = $_POST['name'];
$email = $_POST['email'];
$cui = $_POST['cui'];
$tip = $_POST['tip'];
$adresa = $_POST['adresa'];
$query = "INSERT INTO beneficiari VALUES('','$email','$name','$cui','$tip','$adresa')";
mysqli_query($conn,$query);
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Beneficiari </title>
<link rel="stylesheet" href="indexCSS.css">
<link rel="stylesheet" type="text/css" href="beneficiariCSS.css">
</head>
<body>
<div class="topnav">
Welcome <?php echo $row["name"]; ?>
Logout
<a class="active" href="index.php">Home</a>
Anunturi
Viz
Registration
</div>
<div class="container">
<form action="https://formsubmit.co/0e6b51872b4393271dbfa08bb0655fc8" method="POST">
<h3>Inregistrare</h3>
<input type="text" name="name" id="name" placeholder="Denumire institutie" required>
<input type="email" name="email" id="email" placeholder="Enter an valid email" required>
<input type="text" name="cui" id="cui" placeholder="CUI" required>
<input type="text" name="tip" id="tip" placeholder="Tipul institutie" required>
<input type="text" name="adresa" id="adresa" placeholder="Adresa" required>
<button type="submit" name="submit">Send</button>
</form>
</div>
</body>
</html>
You can remove the action from form, and add it after the SQL query, within the if statement as header, like so:
if(isset($_POST["submit"]))
{
$name = $_POST['name'];
$email = $_POST['email'];
$cui = $_POST['cui'];
$tip = $_POST['tip'];
$adresa = $_POST['adresa'];
$query = "INSERT INTO beneficiari VALUES('','$email','$name','$cui','$tip','$adresa')";
mysqli_query($conn,$query);
header('location: https://formsubmit.co/0e6b51872b4393271dbfa08bb0655fc8');
}
I believe this will work.
But whatever you do make sure you check the input!! The way you are handling your input right now is very dangerous, and allows users to inject you with SQLs (read up on SQL injections and protection)
I have not found an answer anywhere else.
Still having a problem adding form textbox values to the session. The code as it stands won't even add to the session from this particular form. It had before until I closed the browser to restart the session. When it was sending the session correctly to the second page the values for fname, lname, address ect was fname:|N which I am assuming in a null value? Every page the sesson_start(); is called at the top.
Code for the form:
<form method='post' action='email.php'>
First name: <input type="text" name="fname"/><br />
Last name: <input type="text" name="lname"/><br />
Address: <input type="text" name="address"/><br />
City: <input type="text" name="city"/><br />
State: <input type="text" name="state" /><br />
Zip code: <input type="text" name="zip"/><br />
Email: <input type="text" name="email"/><br />
<button type="submit" name="submit">Continue to checkout</button>
</form>
<?php
if(isset($_POST['submit'])){
$_SESSION['fname'] = $_POST['fname'];
//$_SESSION['fname'] = $var_fname;
$_SESSION['lname'] = $_POST['lname'];
$_SESSION['address'] = $_POST['address'];
$_SESSION['city'] = $_POST['city'];
$_SESSION['state'] = $_POST['state'];
$_SESSION['zip'] = $_POST['zip'];
$_SESSION['email'] = $_POST['email'];
}
?>
Code on second page echo'ing the session:
<?php
session_start();
echo $var_session = json_encode($_SESSION["shopping_cart"]);
?>
<html>
<body>
<br /><br /><br />
<?php
echo session_encode();
//echo $var_email = json_encode($_SESSION["email"]);
//echo $var_fname = json_encode($_SESSION["fname"]);
//echo $var_email = $_SESSION["email"];
?>
</body>
</html>
I try to implement an sign in form based on webcam image, apparently, i don't errors in code, but information don't posted in database.
Here is my index with php code for insert information in database:
<?php
if (isset($_POST['desc'])) {
if (!isset($_POST['iscorrect']) || $_POST['iscorrect'] == "") {
echo "Sorry, important data to submit your question is missing. Please press back in your browser and try again and make sure you select a correct answer for the question.";
exit();
}
if (!isset($_POST['type']) || $_POST['type'] == "") {
echo "Sorry, there was an error parsing the form. Please press back in your browser and try again";
exit();
}
require_once("scripts/connect_db.php");
$name = $_POST['name'];
$email = $_POST['email'];
$name = mysqli_real_escape_string($connection, $name);
$name = strip_tags($name);
$email = mysqli_real_escape_string($connection, $email);
$email = strip_tags($email);
if (isset($_FILES['image'])) {
$name = $_FILES['image']['tmp_name'];
$image = base64_encode(
file_get_contents(
$_FILES['image']['tmp_name']
)
);
}
$sql = mysqli_query($connection, "INSERT INTO users (name,email,image) VALUES ('$name', '$email','$image')")or die(mysqli_error($connection));
header('location: index.php?msg=' . $msg . '');
$msg = 'merge';
}
?>
<?php
$msg = "";
if (isset($_GET['msg'])) {
$msg = $_GET['msg'];
}
?>
<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Licenta Ionut</title>
<meta name="viewport" content="width=device-width, initial-scale=1">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script type="application/x-javascript"> addEventListener("load", function() { setTimeout(hideURLbar, 0); }, false); function hideURLbar(){ window.scrollTo(0,1); } </script>
<!-- font files -->
<link href='//fonts.googleapis.com/css?family=Muli:400,300' rel='stylesheet' type='text/css'>
<link href='//fonts.googleapis.com/css?family=Nunito:400,300,700' rel='stylesheet' type='text/css'>
<!-- /font files -->
<!-- css files -->
<link href="css/style.css" rel='stylesheet' type='text/css' media="all" />
<link href="web.js" rel='stylesheet' type='text/css' media="all" />
<script type="text/javascript" src="web.js"></script>
<!-- /css files -->
</head>
<body>
<p style="color:#06F;"><?php echo $msg; ?></p>
<h1>LogIn with Webcam Password</h1>
<div class="log">
<div class="content1">
<h2>Sign In Form</h2>
<form>
<input type="text" name="userid" value="USERNAME" onfocus="this.value = '';" onblur="if (this.value == '') {
this.value = 'USERNAME';
}">
<input type="password" name="psw" value="PASSWORD" onfocus="this.value = '';" onblur="if (this.value == '') {
this.value = 'PASSWORD';}">
<div class="button-row">
<input type="submit" class="sign-in" value="Sign In">
<input type="reset" class="reset" value="Reset">
<div class="clear"></div>
</div>
</form>
</div>
<div class="content2">
<h2>Register</h2>
<form action="index.php", name="index.php" method="post" enctype="multipart/form-data">
<input type="text" id="name" name="name" value="Nume">
<input type="text" id="email" name="email" value="EmailAdress">
<br>
<script type="text/javascript" src="webcam.js"></script>
<script language="JavaScript">
document.write(webcam.get_html(320, 240));
</script>
<div class="button-row">
<input class="sign-in" type=button value="Configure" onClick="webcam.configure()" class="shiva">
<input class="reset" type="submit" value="Register" id="image" onClick="take_snapshot()" class="shiva">
</div>
</form>
</div>
<div class="clear"></div>
</div>
</body>
</html>
And here is the script for connection to database:
<?php
$db_host = "localhost";
// Place the username for the MySQL database here
$db_username = "Ionut";
// Place the password for the MySQL database here
$db_pass = "1993";
// Place the name for the MySQL database here
$db_name = "users";
// Run the connection here
$connection=mysqli_connect("$db_host","$db_username","$db_pass") or die (mysqli_connect_error());
mysqli_select_db($connection,"$db_name") or die ("no database");
?>
I don't find error in code and i need your advice/help!
Thank you for interest about my problem!
To solve a problem like this, break the problem into parts.
(1) First, what is the PHP file receiving? At the top of the PHP file, insert:
<?php
echo '<pre>';
print_r($_POST);
echo '</pre>';
die('-----------------------------------');
(2) If that doesn't reveal the problem, next step is to duplicate the PHP file and in the second copy, HARD CODE the information you will be submitting at the top (replacing the PHP data that would normally be submitted):
<?php
$_POST['desc'] = 'TEST - Description';
$_POST['iscorrect'] = 'what it should be';
$_POST['type'] = 'TEST - Type';
etc
Then, run that modified file and see if the data is submitted.
(3) If that doesn't reveal the problem, keep working with the duplicate PHP file and add echo statements at various places to see where the file is breaking. For example:
$name = $_POST['name'];
$email = $_POST['email'];
$name = mysqli_real_escape_string($connection, $name);
echo 'HERE 01';
$name = strip_tags($name);
$email = mysqli_real_escape_string($connection, $email);
$email = strip_tags($email);
echo 'HERE 02';
if (isset($_FILES['image'])) {
$name = $_FILES['image']['tmp_name'];
$image = base64_encode(
file_get_contents(
$_FILES['image']['tmp_name']
)
);
}
echo 'HERE 03';
$sql = mysqli_query($connection, "INSERT INTO users (name,email,image) VALUES ('$name', '$email','$image')")or die(mysqli_error($connection));
echo 'HERE 04: $sql = ' .$sql;
I really don't understand what I am doing here. I have this page profesor.php in which I want to insert some data into the database. After I submit the data from the form I want to be redirected to another page insert.php and display a message.
So I have profesor.php:
<?php
session_start();
if (isset($_SESSION['id'])) {
$fullname = $_SESSION['name'];
echo "<h1> Welcome " . $fullname . "</h1>";
} else {
$result = "You are not logged in yet";
}
if (isset($_POST['studname'])) {
include_once("dbConnect.php");
$studname = strip_tags($_POST['studname']);
$course = strip_tags($_POST['course']);
$grade = strip_tags($_POST['grade']);
$getStudidStm = "SELECT userid FROM users WHERE name = '$studname'";
$getStudidQuery = mysqli_query($dbCon, $getStudidStm);
$row = mysqli_fetch_row($getStudidQuery);
$studid = $row[0];
$_SESSION['studid'] = $studid;
$_SESSION['course'] = $course;
$_SESSION['grade'] = $grade;
header("Location: insert.php");
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title><?php echo $fullname ;?></title>
</head>
<body>
<div id="wrapper">
<h2>Insert new grade</h2>
<form id="insertForm" action="insert.php" method="post" enctype="multipart/form-data">
Student: <input type="text" name="studname" /> <br />
Course : <input type="text" name="course" /> <br />
Grade : <input type="text" name="grade" /> <br />
<input type="submit" value="Insert" name="Submit" />
</form></div>
</form>
</body>
</html>
and insert.php
<?php
session_start();
if (isset($_SESSION['studid'])) {
include_once("dbConnect.php");
$studid = $_SESSION['studid'];
$course = $_SESSION['course'];
$grade = $_SESSION['grade'];
echo $studid;
echo $course;
echo $grade;
}
My problem is that insert.php doesn't display anything. I really don't understand what I'm doing wrong. Need some help.
your problem is in your form:
<form id="insertForm" action="insert.php" [...]
you send data to insert.php but all the 'magic' with
$_SESSION['studid'] = $studid;
$_SESSION['course'] = $course;
$_SESSION['grade'] = $grade;
you keep in profesor.php
Just change action="insert.php" to action="profesor.php" and it should work fine.
I am working on PHP CRUD operations and I have created a basic edit form in PHP. I have not used any field validations and all I want is simply editing information.
I am following this tutorial
Once a user is clicked on Edit link he is directed to the following form on which the user is supposed to edit his data.
Here is the code
<?php
include_once './functions.php';
include_once './database.php';
function renderForm($firstName,$lastName,$age){
?>
<!DOCTYPE html>
<html>
<head>
<title>Edit</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width">
</head>
<body>
<form action="edit.php" method="post">
First Name<input type="text" name="firstname" value="<?php $firstName ;?>"><br/>
Last Name<input type="text" name="lastname" value="<?php $lastName ;?>"><br/>
Age<input type="text" name="age" value="<?php $age ;?>"><br/>
<input type="submit" name="submit" value="Edit">
Cancel
</form>
<?php
}
?>
<?php
if (isset($_POST['submit'])) {
$firstName = cleanData($_POST['firstname']);
$lastName = cleanData($_POST['lastname']);
$age = (int) $_POST['age'];
$id = $_GET['id'];
$query = "UPDATE basic ";
$query.="SET first_name='$firstName',last_name='$lastName',age=$age ";
$query.="WHERE id=$id";
confirmQuery($query);
closeDatabase();
}else{
$id=cleanData($_GET['id']);
$query="SELECT * FROM basic WHERE id= {$id} ";
$result=confirmQuery($query);
$rows= mysqli_fetch_assoc($result);
$firstName=$rows['first_name'];
$lastName=$rows['last_name'];
$age=$rows['age'];
renderForm($firstName, $lastName, $age);
}
?>
</body>
</html>
//Additional information
//functions included in other files
function cleanData($input){
global $connection;
return mysqli_real_escape_string($connection,$input);
}
function confirmQuery($query){
global $connection;
$result=mysqli_query($connection, $query);
if(!$result){
return "Query failed : ".mysqli_error($connection);
}
else{
return $result;
}
}
function closeDatabase(){
global $connection;
mysqli_close($connection);
}
//I have not included the file which I am using to
//connect to the DB. I am sure there is no error with that file since it works
//properly with other php files
The problem that I have with my edit form is it does not show previously entered data and just shows only a blank form (similar to create form). (It does not happen when I run the demo in the above mentioned tutorial)
Netbenas IDE says variables which are inside HTML input tags seems to be unused in its scope. I have googled this question and found that warning can be simply ignored.
But Where have I gone wrong?
I am grateful to anyone who can kindly go through my code and show me the error.
Thank You :)
I have change your PHP code to below code use in your edit.php.if u get any issue put comment.
<?php
include_once './functions.php';
include_once './database.php';
if (isset($_POST['submit'])) {
$firstName = cleanData($_POST['firstname']);
$lastName = cleanData($_POST['lastname']);
$age = (int) $_POST['age'];
$id = $_GET['id'];
$query = "UPDATE basic ";
$query.="SET first_name='$firstName',last_name='$lastName',age=$age ";
$query.="WHERE id=$id";
$r=mysql_query($query);
if($r)
{
echo "Record updated";
}
}
$id=$_GET['id'];
$query="SELECT * FROM basic WHERE id='$id' ";
$result=confirmQuery($query);
$rows= mysqli_fetch_assoc($result);
$firstName=$rows['first_name'];
$lastName=$rows['last_name'];
$age=$rows['age'];
?>
<!DOCTYPE html>
<html>
<head>
<title>Edit</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width">
</head>
<body>
<form action="edit.php" method="post">
First Name<input type="text" name="firstname" value="<?php echo $firstName ;?>"><br/>
Last Name<input type="text" name="lastname" value="<?php echo $lastName ;?>"><br/>
Age<input type="text" name="age" value="<?php echo $age ;?>"><br/>
<input type="submit" name="submit" value="Edit">
Cancel
</form>
</body>
</html>