How I add the information in my database depending on what is selected.?

Question

I have this select

 <select name="tipi">
          <option value="gp">Gradinita Privata</option>
          <option value="gs">Gradinita de stat</option>
          <option value="cr">Cresa</option>
 </select>

I want to add the information in my database depending on what is selected. I tried

$tip="";
//here is more information
$tip=$_POST["tipi"];

    
   if( $tip === 'gp'){
$sql="INSERT INTO gradiniteprivate (denumire,email,adresa,telefon,pret) VALUES 
('$denumire','$email','$adresa','$telefon','$pret')";
echo $sql;
echo "</br>";}
else if( $tip === 'gs'){
    $sql="INSERT INTO gradinitestat (denumire,email,adresa,telefon,pret) VALUES 
('$denumire','$email','$adresa','$telefon','$pret')";
echo $sql;
echo "</br>";
}
else if( $tip === 'cr'){
      $sql="INSERT INTO crese (denumire,email,adresa,telefon,pret) VALUES 
('$denumire','$email','$adresa','$telefon','$pret')";
echo $sql;
echo "</br>";
}

I have these errors: Warning: Undefined array key "tipi" in C:\xampp\htdocs\Proiect\adaugag-d.php on line 31

Warning: Undefined variable $sql in C:\xampp\htdocs\Proiect\adaugag-d.php on line 49

Fatal error: Uncaught ValueError: mysqli_query(): Argument #2 ($query) cannot be empty in C:\xampp\htdocs\Proiect\adaugag-d.php:49 Stack trace: #0 C:\xampp\htdocs\Proiect\adaugag-d.php(49): mysqli_query(Object(mysqli), '') #1 {main} thrown in C:\xampp\htdocs\Proiect\adaugag-d.php on line 49

There is no particular problem with the code you entered, The error messages you posted are from another section, This question needs to be told in more detail — Erfan Bahramali, May 08 '21 at 20:18
Does the html form and the photos code are in the same file? If yes, you should test that $_POST is initialized with [isset](https://www.php.net/manual/en/function.isset.php) function — DonKnacki, May 08 '21 at 21:04
**Warning:** You are wide open to [SQL Injections](https://php.net/manual/en/security.database.sql-injection.php) and should use parameterized **prepared statements** instead of manually building your queries. They are provided by [PDO](https://php.net/manual/pdo.prepared-statements.php) or by [MySQLi](https://php.net/manual/mysqli.quickstart.prepared-statements.php). Never trust any kind of input! Even when your queries are executed only by trusted users, [you are still in risk of corrupting your data](http://bobby-tables.com/). [Escaping is not enough!](https://stackoverflow.com/q/5741187) — Dharman, May 08 '21 at 21:09

score 1 · Answer 1 · answered May 08 '21 at 20:43

I have these errors: Warning: Undefined array key "tipi" in C:\xampp\htdocs\Proiect\adaugag-d.php on line 31

Your either accessing over GET request and POST is not set or your form is not actually posting the value, <select name="tipi"> outside </form>?

Warning: Undefined variable $sql in C:\xampp\htdocs\Proiect\adaugag-d.php on line 49

It's a cascading error, you are not defining $sql because it's only defined inside an if statement which is not truthy due to previous issue, not POST'ing value, or GET request.

Fatal error: Uncaught ValueError: mysqli_query(): Argument #2 ($query) cannot be empty in C:\xampp\htdocs\Proiect\adaugag-d.php:49 Stack trace: #0 C:\xampp\htdocs\Proiect\adaugag-d.php(49): mysqli_query(Object(mysqli), '') #1 {main} thrown in C:\xampp\htdocs\Proiect\adaugag-d.php on line 49

Again, it's expecting $sql to be set.

You should check for POST request, validate user input and use prepared queries.

<?php
$errors = [];

// its a POST!
if ($_SERVER['REQUEST_METHOD'] === 'POST') {

   // validate form inputs
   if (!isset($_POST["tipi"])) {
     $errors['tipi'] = 'Tipi is a required field';
   } elseif(!in_array($_POST["tipi"], ['gp', 'gs', 'cr'])) {
     $errors['tipi'] = 'Invalid tipi value';
   }

   // @todo: add validation for $denumire, $email, $adresa, $telefon, $pret

   // $errors is empty so must be no errors, continue to do query..
   if (empty($errors)) {

     // determine table, could use an if statement, or not do it as its already been validated as either gp, gs  or cr
     $table = '';
     switch ($_POST["tipi"]) {
      case: "gp": $table = 'gradiniteprivate'; break;
      case: "gs": $table = 'gradinitestat'; break;
      case: "cr": $table = 'crese'; break;
     }
   
     // run prepared query
     $stmt = $mysqli->prepare("
       INSERT INTO $table (
         denumire,
         email,
         adresa,
         telefon,
         pret
       ) VALUES (?,?,?,?,?)");
     $stmt->bind_param("sssss", $denumire, $email, $adresa, $telefon, $pret);
     $stmt->execute();
     
   }
}
?>

// if form is after this
<form>
  ...

  <select name="tipi">
      <option value="gp">Gradinita Privata</option>
      <option value="gs">Gradinita de stat</option>
      <option value="cr">Cresa</option>
  </select>
  <?= !empty($errors['tipi']) ? '<div class="invalid-feedback">'.$errors['tipi'].'</div>' : '' ?>

  ...
</form>

How I add the information in my database depending on what is selected.?

1 Answers1