Hello I have problem with my php code it won't insert value to the database and when it does the value is duplicate.
Here is the php code:
if (isset($_GET['addform']))
{
include '../includes/db.inc.php';
try
{
$sql = 'INSERT INTO author SET Author_name = :Author_name, Author_email =:Author_email';
$s = $pdo->prepare($sql);
$s->bindvalue(':Author_name', $_POST['Author_name']);
$s->bindvalue(':Author_email', $_POST['Author_email']);
$s->execute();
}
catch (PDOException $e)
{
$error = 'Error adding submitted author.';
include 'error.html.php';
exit();
}
header ('Location COMP1321/recipes/admin/authors/authors.html.php');
exit();
}
And here is the html form
<? php include 'index.php' ?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title><?php html($pageTitle); ?></title>
</head>
<body>
<h1><?php html($pageTitle); ?></h1>
<form action="?addform" method="GET">
<label for="name"> Name: <input type="text" name="Author_name" id="Author_name"></label>
<br/>
<label for="email"> Email: <input type="text" name="Author_email" id="Author_email" ></label>
<br/>
<input type="hidden" name="id" value="<?php html($id); ?>">
<input type="submit" value="<?php html($button); ?>">
</form>
</body>
</html>
Any idea what went wrong here?
In your HTML code:
<form action="?addform" method="GET">
You are using GET as form method
And in your PHP code:
$s->bindvalue(':Author_name', $_POST['Author_name']);
$s->bindvalue(':Author_email', $_POST['Author_email']);
you are using POST
Try using same method to submit form from html code and get value in PHP code
Related
I tried to execute the following html code:
<html>
<head>
<title>Listing 9.1 A simple HTML form</title>
</head>
<body>
<form action="listing.php" method="POST">
<p><strong>Name:</strong><br>
<input type="text" name="user"></p>
<p><strong>Address:</strong><br>
<textarea name="address" rows="5" cols="40"></textarea></p>
<p><input type="submit" value="send"></p>
</form>
</body>
</html>
And this is the code of the associated php file (listing.php):
<html>
<head>
<title>Listing Reading input from a form </title>
</head>
<body>
Welcome <?php echo $_POST["user"]; ?><br>
Your email address is: <?php echo $_POST["address"]; ?>
</body>
</html>
I was able to get the form and enter values as shown below:
Form Input
But, when I clicked 'Send Message', it displays only:
Welcome
Your email address is:
Without the values that I entered through the form.
When I tried to run the php file directly from the local host (http://localhost/listing.php), I received these error messages:
Welcome
Notice: Undefined index: user in C:\xampp\htdocs\listing.php on line 7
Your email address is:
Notice: Undefined index: address in C:\xampp\htdocs\listing.php on line 8
I even modified the php code as follows, but still got the same output:
<html>
<head>
<title>Listing Reading input from a form </title>
</head>
<body>
Welcome <?php
if(isset($_POST['submit'];)) {
session_start();
$user = $_POST['user'];
echo "$text";}else {echo 'Could not load text!';}?><br>
Your email address is: <?php echo $_POST["address"]; ?>
</body>
</html>
I would really appreciate it if you could give some advice to make it work.
Thanks
if(isset($_POST['submit'];)) should be changed,so you check if address and user is isset. Furthermore you normally don't want ; in your if statements.
Here i have a optimized version for you. the !empty is added so we also check if the inputs are not empty.
if (isset($_POST["name"] && isset($_POST["address"])) {
if (!empty($_POST["name"] && !empty($_POST["address"]) {
// execute code as u would
}
}
Notice: Undefined index: address in C:\xampp\htdocs\listing.php on line 8
This is caused by the use of the ';' in the if condition.
Remove that and you should be good. for now.
Try this code:
<html>
<head>
<title>Listing Reading input from a form </title>
</head>
<body>
Welcome <?php
if(isset($_POST['submit']))
{
session_start();
$user = $_POST['user'];
echo $user;
echo "<br><br>Your Email Address is: ".$_POST['address'];
}
else
{
echo 'Could not load text!';
}
?>
<br>
<form action="" method="POST">
<p><strong>Name:</strong><br>
<input type="text" name="user"></p>
<p><strong>Address:</strong><br>
<textarea name="address" rows="5" cols="40"></textarea></p>
<p><input type="submit" value="send" name="submit"></p>
</form>
</body>
</html>
In your second code block $text isn't defined anywhere. Do you mean to have $user?
<html>
<head>
<title>Listing Reading input from a form</title>
</head>
<body>
Welcome
<?php
if(isset($_POST['submit'])) {
session_start();
echo isset($_POST['user']) ? $_POST['user'] : "User";
} else {
echo "User";
}
?>
<br>
Your email address is:
<?php
echo isset($_POST["address"]) ? $_POST["address"] : "Address"
?>
</body>
</html>
Try this and let me know:
<?php
session_start();
if(isset($_POST['submit'])) {
$user = $_POST['user'];
echo "Welcome ".$user;
echo "<br>";
echo "Your email address is: ".$_POST["address"];
}else {
// you may get rid of this block if you like
echo 'Could not load text!';
}
?>
<html>
<head>
<title>Listing 9.1 A simple HTML form</title>
</head>
<body>
<form action="" method="POST">
<p><strong>Name:</strong><br>
<input type="text" name="user"></p>
<p><strong>Address:</strong><br>
<textarea name="address" rows="5" cols="40"></textarea></p>
<p><input type="submit" name='submit' value="send"></p>
</form>
</body>
</html>
I am currently creating an HTML form that has 2 fields; name and an address. It also has a way of selecting one of 2 options. The form will either be used to look up the address of a person. In this case, only the name field is required to be filled out. The other option is to add a person to the file. In this case both fields need to be filled out. For some reason, I am not able to get the values that inputted from the form into my PHP file. Please help.
Here is my HTML Form
<html>
<head>
<title> Form </title>
</head>
<body>
<form action="action_page.php" method="post">
<div>
<label for="name">Name: </label>
<input id="name" type="text" name="name"><br>
</div>
<div>
<label for=address">Address: </label>
<input id="address" type="text" name="address"><br>
<input type="radio" name="action" value="lookup">Lookup<br>
<input type="radio" name="action" value="add">Add<br>
<input type="submit" name="submit"><br>
</form>
</body>
</html>
Here is my PHP file
<html>
<head>
<title> PHP </title>
</head>
<body>
<?php
$name = $_POST['name'];
echo "<p>",$_POST["name"],"</p>";
echo "<p>",$_POST["action"],"</p>";
echo "<p>",$_POST["address"],"</p>";
$address = array();
if($_POST["action"]=="lookup"){
$fh = fopen("address.txt","r") or die("No such file found.");
while(!feof($fh)) {
$line = fgets($fh);
$info = explode("|",$line);
$address[$info[0]]=$info[1];
}
if(array_key_exists($_POST["name"],$address)) {
echo "<p>",$_POST["name"],"<p>";
echo "<p>",$address[$_POST["name"]],"</p>";
}
?>
<body>
</html>
The error was in
echo "<p>",$_POST["name"],"</p>";
It should be
echo "<p>".$_POST["name"]."</p>";
and same for others
I have problem in this code.
In this code when i press save data button , the data insert into database but when i refresh page then it's automatically insert into database, what should i do in my code then stop automatically insert data into database, thanks in advance
<?php
require './database/databaseConnection.php';
if(isset($_POST['save_button']))
{
if(($_POST['fname']&&($_POST['lname'])))
{
$first_name=$_POST['fname'];
$last_name=$_POST['lname'];
$qry="INSERT INTO user_master(first_name,last_name) values('$first_name','$last_name')";
$result= mysql_query($qry)or die(mysql_error());
if($result){
echo 'SuccessFully saved data';
}
else{
echo 'Data Not Inserted!';
}
}
}
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title></title>
<link rel="stylesheet" href="bootStrap/css/bootstrap.min.css">
<link rel="stylesheet" href="bootStrap/css/bootstrap.css">
</head>
<body>
<div class="container jumbotron ">
<form action="" method="post">
<table class="table-responsive">
<div class="form-group form-inline">
<tbody>
<tr>
<td> <label for="fname" class="label-info">First Name</label></td>
<td><input class="form-control" type="text" name="fname"></td>
<td><label for="lname" class="label-info">Last Name</label></td>
<td><input class="form-control" type="text" name="lname"></td>
</tr>
</tbody>
</div>
</table>
<button type="submit" name="save_button" class="btn-success" >Save Data</button>
</form>
</div>
</body>
</html>
This is happening because your action is empty
Update your action to this
action="<?php echo $_SERVER['PHP_SELF']; ?>"
Make a separate php file that will insert data to database. Give this in the form action attribute.
<form action="insert.php" method="post">
......
......
</form>
insert.php file
<?php
require './database/databaseConnection.php';
if(isset($_POST['save_button']))
{
if(($_POST['fname']&&($_POST['lname'])))
{
$first_name=$_POST['fname'];
$last_name=$_POST['lname'];
$qry="INSERT INTO user_master(first_name,last_name) values('$first_name','$last_name')";
$result= mysqli_query($qry)or die(mysql_error());
if($result){
echo 'SuccessFully saved data';
}
else{
echo 'Data Not Inserted!';
}
}
}
?>
You can use header() to redirect to your previous page if you want. Thus not allowing the refreshing of insert.php
header("location: your_page.php");
it will be safe if you use Prepared Statements
Take a look
I have two php files. index.php and insert.php. I am inserting some data by using form but I also want to store the index.php url into mysql by using html form. Because I have many pages like index.php which are calling insert.php. Actually when some page call insert.php it should take the url of that page by using form so that i can store that url into database. Code is given below.
<?php
$con=mysqli_connect("localhost","root","","commentdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Everything is working fine but not taking the url
$sql="INSERT INTO comment (id, name, email,comment,url)
VALUES
(null, '$_POST[name]','$_POST[email]','$_POST[comment]', $_POST[url])";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
HTML
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>My Comment Box</title>
</head>
<body>
<?php
//getURL function to get URL
function getURL()
{
/* First check if page is http or https */
//$whichprotocol = $_SERVER['HTTPS'] == 'on' ? 'https' : 'http';
/* Combine different pieces of $_SERVER variable to return current URL */
//return $whichprotocol.'://'.$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
return '://'.$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
}
/*Calling getURL function to display current page URL*/
?>
<form action="insert.php" method="post">
<p>Name:
<input name="name" type="text" />
</p>
<p>Email:
<input name="email" type="text" />
</p>
<p>Comment:</p>
<p>
<textarea name="comment" cols="50" rows="10"></textarea>
</p>
<p>
<--! getURL(); gives me error. It takes url but don't send it insert query -->
<input type="hidden" name="url" id="url" value="<?php getURL();?>" />
</p>
<p>
<input type="submit" name="submit" id="submit" value="Submit" />
</p>
</form>
<br />
<?php
echo getURL();
?>
</body>
</html>
You need to echo the value from the function -
value="<?php echo getURL();?>" />
I'm trying to pass a dynamic value, an ID of a product. Through a form so I can write in a mysql database.
My problem is that the value is not being passed on through the form. I suspect it's due to that I'm trying to pass a value from one table to another table. Should I set a temporary variable for this $item_id instead? Or is it possible to pass anyway? Please help..
My shopping cart, here I echoes out the $item_id, which shows the last $item_id visible on the browser:
<div>
<?php
echo $item_id
?>
<form method="post" action="checkoutpage.php">
<input type="hidden" name="product_bought" value="<?php echo $item_id ?>">
<input type="submit" value="Submit">
</form>
checkoutpage.php:
<?php
session_start(); // Start session first thing in script
// Script Error Reporting
error_reporting(E_ALL);
ini_set('display_errors', '1');
// Connect to the MySQL database
include "storescripts/connect_to_mysql.php";
?>
<!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" />
<title>checkout form</title>
</head>
<body>
<form action="insert.php" method="post">
Name: <input type="text" name="name">
Email: <input type="text" name="email"><br />
Credit Card: <input type="text" name="credit_card">
<input type="hidden" name="product_bought" value='product_bought'>
<input type="submit">
</form>
</body>
</html>
insert.php
<?php
$con=mysqli_connect("localhost","xxxx","xxxx","mystore");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="INSERT INTO orders (name, email, credit_card, product_bought)
VALUES
('$_POST[name]','$_POST[email]','$_POST[credit_card]','$_POST[product_bought]')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
on checkout page you have:
<input type="hidden" name="product_bought" value='product_bought'>
and you should have:
<input type="hidden" name="product_bought" value='<?php echo $_POST['product_bought']; ?>'>
You are passing "product_bought" as the value.
I hope it fixes it.