PHP pass an ID between scripts - php

Question

I have a problem that I can't solve.
I am creating an Android App where there is a section showing user data.
The data is saved on a MySql database through Register.php and to get them use the Show.php file.
When I open the user activity in the app, there is a hidden field that contains the user's id, the app sends the id to the Php script and shows all the data of that user.
The problem is that the data is shown with a bit of delay, because it has to read the id from the app and show the filtered data.
What I would like is that in the Php file there was already the user id, so the display of the data on the app would be instantaneous, because so I already have the value of the variable that contains the id to use then in the Query.
To do this I thought of using sessions.
But it does not work.
In the Register.php file I inserted a session variable to store the value of the variable id_restaurants that I need in the show.php script to show that user's data through a Query.
But in the Register.php file it seems that the value of the variable $ _SESSION ['varId'] is not saved.
What am I doing wrong?
Here is the registration file: Register.php
<?php
session_start();
if($_SERVER['REQUEST_METHOD']=='POST'){
include_once("config2.php");
$idUser = $_POST['idUser'];
$name = $_POST['name'];
$surname = $_POST['surname'];
$restaurant = $_POST['restaurant'];
$id_restaurant = $_POST['id_restaurant'];
$_SESSION['varId'] = '$id_restaurant';
if($name == '' || $surname == '' || $restaurant =='' ){
echo json_encode(array( "statusr" => "false","message" => "Insert all data") );
}else {
$query= "SELECT * FROM Restaurants WHERE restaurant='$restaurant'";
$result= mysqli_query($con, $query);
if(mysqli_num_rows($result) > 0){
echo json_encode(array( "statusr" => "false","message" => "Name restaurant already exist") );
}else{
$query = "INSERT INTO Restaurants (idUser,name,surname,restaurant,id_restaurant) VALUES ('$idUser','$name','$surname','$restaurant','$id_restaurant')";
if(mysqli_query($con,$query)){
$query= "SELECT * FROM Restaurants WHERE name='$name' AND surname='$surname' AND restaurant='$restaurant' ";
$result= mysqli_query($con, $query);
$emparray = array();
if(mysqli_num_rows($result) > 0){
while ($row = mysqli_fetch_assoc($result)) {
$emparray[] = $row;
}
}
echo json_encode(array( "statusr" => "truer","message" => "Registered succesfully!" , "datar" => $emparray) );
}else{
echo json_encode(array( "statusr" => "false","message" => "Error") );
}
}
mysqli_close($con);
}
} else{
echo json_encode(array( "statusr" => "false","message" => "Errore3") );
}
?>
Here is the show file: Show.php
<?php
session_start();
include 'config.php';
$conn = new mysqli($HostName, $HostUser, $HostPass, $DatabaseName);
$idRestaurant = $_SESSION['varId'];
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM Food WHERE id_restaurant ='$idRestaurant'" ;
$result = $conn->query($sql);
if ($result->num_rows >0) {
while($row[] = $result->fetch_assoc()) {
$tem = $row;
$json = json_encode($tem);
}
} else {
echo "No Results Found.";
}
echo $json;
$conn->close();
?>
And then you think there's a better way to pass the id between Php scripts?
I listen to you willingly.
Thanks a lot.