从php中的另一个页面调用函数

I have a form that user can update their details on page-1.php.

After user submit the form, it will got to page-2.php for validation and update database if user input is true.

May i know how to call a function from page-2.php and display the error message on page-1.php. Below are the function on page-2.php that I use to display the error message:

$_SESSION['err_msg'] = array();
function addError($msg) {   
    $_SESSION['err_msg'][] = $msg;
}

function printErrors() {
    foreach ($_SESSION['err_msg'] as $err){
        echo "<ul><li><span style='color: red;'>".$err."</span></li></ul>";
    }
    unset($_SESSION['err_msg']);
}   
//other codes for the validation and update

if i use printErrors();on page-2.php, it will display the error message

You need to include page-2.php in page-1.php. Read this about that.

The best way is to put all your functions in one seperate file, like functions.php and include them on every page with an include.

I think no you don't need function on page-2.php. You can simply check session variable on page-1.php. If $_SESSION['err_msg'] exists and count is greater than 0. Then you can print all errors to page-1.php> Please check below code.

if(isset($_SESSION['err_msg']) && count($_SESSION['err_msg'])>0 ){
foreach ($_SESSION['err_msg'] as $err){
    echo "<ul><li><span style='color: red;'>".$err."</span></li>    </ul>";
}
unset($_SESSION['err_msg']);
}

Hope this helps you.

Include the page-2.php in your page-1.php