从另一个外部php脚本运行外部php脚本

Question

i have my main index page which uses an ajax request to an external php file which handles the user login data. In that external php file i also include another external php file which handles all the functions, however the external login file is not able to use any of the functions

Here is the ajax call from index.php

$('#ind_login_submit').on('click', function (e) {

    var vars = $("#ind_login_form").serialize(); // the script where you handle the form input.
    //alert("gu");
    var hr = new XMLHttpRequest();
    hr.open("POST", "scripts/index/ind_login_submit.php", true);
    hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    hr.onreadystatechange = function() {
        if(hr.readyState == 4 && hr.status == 200) {
            var data = JSON.parse(hr.responseText);
            for(var obj in data){
                if(obj == "error"){
                    alert(data[obj]);

                }else if(obj == "success"){
                    alert(data[obj]);
                    window.location.replace("http://localhost/site/dashboard.php");
                }
            }
            //alert(hr.responseText);
            //location.load();
        }
    };
    hr.send(vars);
    //results.innerHTML = "requesting...";
    event.preventDefault();
});

Here is the external ind_login_submit.php

    header("Content-Type: application/json");
 session_start();
   include '../../connect.php';
    include '../functions.php';
    $secret_key = '';
globalSecret($secret_key);
$error_array = array('error' => $secret_key);
            $jsonData = json_encode($error_array);
            echo $jsonData; 
            exit;

if(isset($_POST['ind_login_remember'])){

    $ind_login_remember=1;
}else{

    $ind_login_remember=0;
}



$ind_login_email = $_POST['ind_login_email'];
$ind_login_password = $_POST['ind_login_password'];

And here is the functions.php

function globalSecret(){

$secret = "This is the secret";
$secret_key = sha1($secret);
}

When i run the code i just get a blank alert show, where it should show the $secret_key variable

Answer 1

I think the line

globalSecret($secret_key);

should be

$secret_key = globalSecret();

and the function globalSecret() should be as follows:

function globalSecret(){
    $secret = "This is the secret";
    $secret_key = sha1($secret);
    return $secret_key;
}