php: Execute php file from another php file with parameters - php

Question

Details: I have a php file that runs some sql and retrieve a list of information. I run through a loop of that information and want to call another php page and pass it some parameters from the data I am looping through.
Question: How do I execute another php page from within my php?
What I Have Tried: This is the php code that should be calling the second php page (the while loop should be calling the simplepush.php page for each result I get):
<?php
require_once "../database/config.php";
header("Content-type: application/json");
$sql = "SELECT user_ip_address FROM ft_users";
$res = mssql_query($sql);
if (mssql_num_rows($res)) {
while ($op = mssql_fetch_assoc($res)) {
exec('simplepush.php?token = ' . $op . '');
$arr[] = $op;
}
echo json_encode($arr);
//$op = mssql_fetch_assoc($res);
//$op['response'] = 200;
} else {
http_response_code(420);
$op = array(
'response' => 420
);
echo json_encode($op);
}
mssql_close();
?>
I have tried the following:
include ('simplepush.php?token = '.$op.'');
exec ('simplepush.php?token = '.$op.'');
require ('simplepush.php?token = '.$op.'');
shell_exec ('simplepush.php?token = '.$op.'');

Answer 1

Just do:
include ('simplepush.php');
Now $op will be available in simplepush.php. Consider this example:
//index.php
while ($op = mssql_fetch_assoc($res)) {
include ('simplepush.php');
$arr[] = $op;
}
//simplepush.php
print_r($op);
The contents of $op will be output each time through the loop.