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How to pass values of select query to the next page inside variable?


shams

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Hi,

The index.php calling the check.php  with the form action="check.php" and passing the value of $cat_id and $subcat to the check .php from the two drop down menu,  i i included the other  page post.php with require_once in the check.php, the post.php gets the value of $subcat from the index.php as condition  and prints the full row of msqyl table subcategory, my question is how to pass the values of post.php in the variable $_POST to the check.php as you see below it prints the row  but not with the $_POST variables, this is the output of all pages in the check.php:

cat_id =4 subcategory=user@gmail.com user=user@gmail.com passw=aaaaaaa server=smtp.gmail.com



Notice: Undefined index: user in /home/user/www/user.be.eu.org/check.php on line 5

Notice: Undefined index: passw in /home/user/www/user.be.eu.org/check.php on line 6

Notice: Undefined index: server in /home/user/www/user.be.eu.org/check.php on line 7
Value of $cat = 4
Value of $subcat = user@gmail.com
Value of $user =
Value of $passw =
Value of $server = 

This is check.php:

 

<?Php
require_once 'post.php';
$cat= $_POST['cat'];
$subcat= $_POST['subcat'];
$from = $_POST['from'];
$user = $_POST['user'];
$passw = $_POST['passw'];
$server = $_POST['server'];
print_r($row);	
	
echo "Value of \$cat = $cat <br>Value of \$subcat = $subcat <br> Value of \$from = $from <br>Value of \$user = $user <br>Value of \$passw = $passw <br>Value of \$server = $server";
?>

  this is post.php:

<?php
set_exception_handler(function($e) {
  error_log($e->getMessage());
  exit('Something weird happened'); //something a user can understand
});
 error_reporting(E_ALL);

        ini_set('display_errors', '1');



$host = 'localhost';
$db   = 'mail';
$user = 'root';
$pass = 'root';
$charset = 'utf8mb4';

$options = [
    PDO::ATTR_ERRMODE            => PDO::ERRMODE_EXCEPTION,
    PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC,
    PDO::ATTR_EMULATE_PREPARES   => false,
];
$pdo = new PDO("mysql:host=$host;dbname=$db;charset=$charset", $user, $pass, $options);
/////
session_start();
$subcat=$_POST['subcat'];
// select a particular user by id
$stmt = $pdo->prepare("SELECT * FROM subcategory  WHERE subcategory=?");
$stmt->execute([$subcat]); 
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
    echo  ('cat_id =' . $row['cat_id'] . "\r\n" . 'subcategory=' . $row['subcategory']. "\r\n". 'user=' . $row['user'] . "\r\n". 'passw=' . $row['passw']. "\r\n". 'server=' . $row['server']  . '<br />');

}

?>
<!DOCTYPE html>
<html>
	<head></head>
	<body>
		<br />
		<div class="container">
			<div class="row">
				<div class="col-md-8" style="margin:0 auto; float:none;">
						<br />
                  <form action = "" method = 'post'>
                  <input type="hidden" name="from" value="<?php echo $_POST['subcategory']; ?>" />
                  <input type="hidden" name="user" value="<?php echo $_POST['user']; ?>" />
                  <input type="hidden" name="passw" value="<?php echo $_POST['passw']; ?>" />
                  <input type="hidden" name="server" value="<?php echo $_POST['server']; ?>" />
               	</form>
				</div>   
			</div>
		</div>
	</body>
</html> 

 

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Problem solved with  this  modified  post.php:

$pdo = new PDO("mysql:host=$host;dbname=$db;charset=$charset", $user, $pass, $options);
/////
session_start();
$subcat=$_POST['subcat'];
$stmt = $pdo->prepare("SELECT * FROM subcategory  WHERE subcategory=?");
$stmt->execute([$subcat]); 
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
  $cat = $row['cat_id'];
  $from = $row['subcategory'];
  $user = $row['user'];
  $passw= $row['passw'];
  $server = $row['server']; 
}

 

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