Show records in the table with if - PHP - php

Question

I have a plant table and a customer campaign table, I want to show only plants that are not associated with the customer campaign table.
through this if cycle, I would like to get a solution like if it is true not to show the plants that are associated if and it is false shows the plants that are not associated
it's possible?
<?php
session_start();
include 'connessione.php';
$var = true;
$var = 1;
$query = mysqli_query($connessione, "
SELECT *
FROM store_locator
INNER JOIN campagne_cliente
ON store_locator.id = campagne_cliente.impianto_id_campagna");
if (!$query)
{
die('Error: ' . mysqli_error($connessione));
}
if($var === true){
echo "email already exists";
}else{
echo "ok";
}
?>
but if i have only plants associated and nothing plants free i have always ok.
Why?

Answer 1

Correct:
<?php
session_start();
include 'connessione.php';
$id = $_SESSION['id'];
$query_string = "SELECT * FROM store_locator WHERE store_locator.id NOT IN (SELECT impianto_id_campagna FROM campagne_cliente);
";
$query = mysqli_query($connessione, $query_string);
?>
<?php
while($row = mysqli_fetch_assoc($query)){ ?>
<?php echo $row['id'] ;?>
<?php } ?>

Answer 2

USE NOT IN QUERY LIKE THIS....
AM Just Try to understand you,you can manage it according to your table structure.
SELECT * FROM `plant_table` not in (select customer_campaign.plant_id from customer_campaign);
Hope This will Help You.
Thanks
I got this result:
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'not in (select campagne_cliente.impianto_id_campagna from campagne_cliente)' at line 2
this is a query:
SELECT * FROM `store_locator`
not in
(select campagne_cliente.impianto_id_campagna
from campagne_cliente);