Added database rows are empty mysql/php - php

Question

I want to write from my form to my database. I'm confused because this resembles the scripts from tutorials and there it works.
Form (w3schools example) extract:
<form action="insert.php" method="post">
Firstname: <input type="text" name="firstname">
Lastname: <input type="text" name="lastname">
Age: <input type="text" name="age">
<input type="submit">
</form>
php:
<?php
$con=mysqli_connect("localhost","XXX","AAA","databasename");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// escape variables for security
$firstname = mysqli_real_escape_string($_POST['firstname']);
$lastname = mysqli_real_escape_string($_POST['lastname']);
$age = mysqli_real_escape_string($_POST['age']);
$sql="INSERT INTO test (firstname, lastname, age)
VALUES ('$firstname', '$lastname', '$age')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
This adds a new row to my database with each submission. The problem: this added row is empty, except for the age column which is always 0, regardless of what I submit.
Where is my mistake?

Answer 1

Refer to php document you must give two values to mysqli_real_escape_string.
try this:
$firstname = mysqli_real_escape_string($con, $_POST['firstname']);
$lastname = mysqli_real_escape_string($con, $_POST['lastname']);
$age = mysqli_real_escape_string($con, $_POST['age']);