How to send data to MySQL using PHP?

Question

PHP

Colleagues, I don’t understand why the data is not added to the database?
Submission Form

<meta charset="UTF-8">
<form method="post" action="action.php">
  Адрес: <input size="255" name="address" type="text">
  IP-сервера: <input size="255" name="IP" type="text">
  <input value="Добавить запись" type="submit" name="submit">
  </form>

Processing script

<?php header('Content-Type: text/html; charset=utf-8');
 include ('db.php');
 error_reporting (E_ALL);
 if(isset($_POST['submit']))

 $address = $_POST['address']; 
 $IP = $_POST['IP']; 
 $sql = 'INSERT INTO table(address, IP, Login, TV, Terminal, monitoring)
 VALUES("'.$address.'", "'.$IP.'", "'.$IP.'", "'.$IP.'", "'.$IP.'", "'.$IP.'")';
// проверка
 if(!mysql_query($sql))
 {echo '<center><p><b>Ошибка при добавлении данных!</b></p></center>';}
 else
 {echo '<center><p><b>Данные добавлены!</b></p></center>';}
?>

only this on the screen:
Connection to the adminka database succeeded.
Error adding data!
Doesn't give any errors. Dots may be present in the IP field, but as far as I remember, this is not critical?

Answer 1

show DB schema.
well, use mysql_* - mauvais ton.

Answer 2

The MySQL server returns an error message to the interpreter - figure out how to display it. Use the power of PDO . Check it out, it 's worth it .
Among other things,

if(isset($_POST['submit']))        // у Вас по данному условию выполняется только
   $address = $_POST['address'];   // это

// А вот всё следующее выполняется В ЛЮБОМ СЛУЧАЕ:

$IP = $_POST['IP'];
$sql = 'INSERT INTO table(address, IP, Login, TV, Terminal, monitoring) '
    . 'VALUES("'.$address.'", "'.$IP.'", "'.$IP.'", "'.$IP.'", "'.$IP.'", "'.$IP.'")';

// проверка
if(!mysql_query($sql)) {
    echo '<center><p><b>Ошибка при добавлении данных!</b></p></center>';
} else {
    echo '<center><p><b>Данные добавлены!</b></p></center>';
}

Answer 3

1. Print the contents of $sql, execute in PhpMyAdmin, carefully read what is the error
2. Do not use mysql_* functions, they are obsolete, and the latest version of PHP does not exist at all
3. Do not substitute data directly, you need EXCLUSIVELY through prepared expressions