$_POST method showing undefine variable - php

Question

How to solve this i don't add data using by $_Post Method, but without $_POST method showing undefine variable?
if (isset($_POST['firstname']) && isset($_POST['lastname']) && isset($_POST['age'])){
//Getting values
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$age = $_POST['age'];
//Creating an sql query
$sql = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";
//Importing our db connection script
require_once('connect.php');
//Executing query to database
if(mysqli_query($con,$sql)){
echo 'Employee Added Successfully';
}else{
echo 'Could Not Add Employee';
}
//Closing the database
mysqli_close($con);
}

Answer 1

You need to define your variables to something if $_POST is not set.
So code would look something like this:
$firstname = isset($_POST['firstname']) ? $_POST['firstname'] ? "";
$lastname = isset($_POST['lastname']) ? $_POST['lastname'] : "";
$age = isset($_POST['age']) ? $_POST['age'] : "";
//if isset $_POST['age'] then assign it to $_POST['age'] else assign it to ""
if (isset($_POST['firstname']) && isset($_POST['lastname']) && isset($_POST['age'])){
//Creating an sql query
$sql = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";
//Importing our db connection script
require_once('connect.php');
//Executing query to database
if(mysqli_query($con,$sql)){
echo 'Employee Added Successfully';
}else{
echo 'Could Not Add Employee';
}
//Closing the database
mysqli_close($con);
}
Bonus:
You code can be easily injected. Use prepared statements to avoid this.
$sql = "INSERT INTO info (firstname,lastname,age) VALUES (?,?,?)"; // question marks are placeholders to bind values to
$stmt = $con->prepare($sql); // prepare query
$stmt->bind_param("sss", $firstname, $lastname, $age); // bind values to your query
if($stmt->execute()){ // if success
echo 'Employee Added Successfully';
}else{
echo 'Could Not Add Employee';
}
You can learn more about it here

Answer 2

Try changing this one:
$sql = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";
To this one:
$sql = "INSERT INTO info (firstname,lastname,age) VALUES ('".$firstname."','".$lastname."','."$age."')";