每次用户登录时（或会话处于活动状态时）如何调用脚本

Question

I am trying run a script called random_post_generator.php which should execute every time a user is logged in.

I am using this approach as an alternative to cron.

Here is how my session is currently created:

<?php
ob_start();
session_start();
if (!isset($_SESSION["user_login"])) {
    header("Location: index.php");
} else {
    $username = $_SESSION["user_login"];
}
?>

But how do I say - "if session is active, then run this script"?

Answer 1

<?php
ob_start();
session_start();
if (!isset($_SESSION["user_login"])) {
    header("Location: index.php");
} else {
    $username = $_SESSION["user_login"];
    include 'random_post_generator.php';
}
?>

or you can use require 'random_post_generator.php'

Answer 2

If I understood correctly, you are trying to find out how to include a script of php (that is located in an outside .php file) inside your current file while using your previous code that checks if a user is logged in:

<?php
$root_directory_path = $_SERVER['DOCUMENT_ROOT'];
ob_start();
session_start();
if (!isset($_SESSION["user_login"])) {
    header("Location: index.php");
} else {
    $username = $_SESSION["user_login"];
    $pathName = $root_directory_path."myScript.php";//I am assuming here
    //the script is located inside the root directory, and not in a sub
    //directory
    require($pathName);
}
?>

just remember that whatever php code is inside myScript.php has to have the <?php ?> tag surrounding it. Your code does not reuse the <?php ?> tag of the "calling" file.

Let me know if that worked for you