PHP SQL prepared statements

[johannes12]

Hallo,
Ich würde gerne Daten aus einem Kontaktformular in eine Tabelle übermitteln. Hier der Code, den ich so weit verfasst habe:

$servername = "";
$username = "";
$password = "";
$dbname = "";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);

// Check connection
if (!$conn) {

    die("Connection failed: " . mysqli_connect_error());
}

$sqlid = "SELECT id FROM contact ORDER BY id DESC LIMIT 1";
$result = mysqli_query($conn, $sqlid);
$row=mysqli_fetch_assoc($result);
$last_id = $row["id"];

if ($last_id == "") {
$fromid = "0"; global $fromid;
} else {
$fromid = $last_id+1; global $fromid;
}

global $fromid;
global $fromname, $frommail, $name_company, $mail_company, $phone_company, $adress_company, $city_company, $country_company, $text;

$stmt = mysqli_stmt_init($conn);
$sql = mysqli_stmt_prepare($stmt, 'INSERT INTO contact (id, name, mail, company, mail_company, phone_company, adress_company, city, country, message)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?)');
mysqli_stmt_bind_param($sql, 'isssssssss', $fromid, $fromname, $frommail, $name_company, $mail_company, $phone_company, $adress_company, $city_company, $country_company, $text);
mysqli_stmt_execute($stmt);
mysqli_stmt_error($stmt); //debugging

print_r($_POST); //debugging

/* $sql = "INSERT INTO contact (id, name, mail, company, mail_company, phone_company, adress_company, city, country, message)
VALUES ('$fromid', '$fromname', '$frommail', '$name_company', '$mail_company', '$phone_company', '$adress_company', '$city_company', '$country_company', '$text')"; */

if (mysqli_query($conn, $sql)) {
    echo "New record created successfully";

} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_stmt_close($stmt);
mysqli_close($conn); 

Ich erhalte folgende Fehlermeldung:

Warning: mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, boolean given in C:\xampp\htdocs\websitetemplates\white_blue_scroll bar_1200px_width\content\connect.php on line 38

Folgendermaßen sieht die Struktur der Tabelle 'user' aus:


Danke im Voraus.

edit:
Achja, $fromname, $frommail, $name_company, $mail_company, $phone_company, $adress_company, $city_company, $country_company und $text wurden außerhalb der Funktion definiert, die hier im Code Snippet zu sehen ist, deswegen sind diese global. All diese Variablen sind folgendermaßen definiert worden:

$fromname = $_POST["fromname"]; 

[h3ll]

https://www.php.de/forum/webentwickl...red-statements

PHP SQL prepared statements - PHP Forum: phpforum.de