PHP脚本在线工作，但没有脱机

This script works online but not offline via WAMP...

function current_user()
    {
        static $current_user;
        if(!$current_user)
        {
            if($_SESSION['userID'])
            {
                $userID = intval($_SESSION['userID']);
                $query = "SELECT * 
                          FROM `users`
                          WHERE `id` = $userID";
                $data  = mysql_query($query);
                if(mysql_num_rows($data))
                {
                    $current_user = mysql_fetch_assoc($data);
                    return $current_user;
                }
            }
        }
        return $current_user;
    }

When this function is called, I get the following error message...

Notice: Undefined index: userID in C:\wamp\www\alpha\_includes\session.php on line 38

Line 38 is $userID = intval($_SESSION['userID']);

The error only occurs when the userID SESSION variable isn't set. That and when the function is called via a page from my offline testing computer (running WAMP).

As a side note, this isn't my script. Credit goes to Jim Hoskins of Think Vitamin Membership.

You have notices enabled on one server, but not the other. This notice is just telling you that are attempting to read an array element that doesn't exist.

Instead, try this

if(isset($_SESSION['userID']))

It's a notice, not an error. To fix:

if (isset($_SESSION['userID']) && $_SESSION['userID'])

change if($_SESSION['userID']) to if(isset($_SESSION['userID'])) first check whether the session variable 'userID' is set before we use it.

Try this code

function current_user()
    {
        static $current_user;
        if(!$current_user)
        {
            if($_SESSION['userID'])
            {
                $userID = intval($_SESSION['userID']);
                $query = "SELECT * 
                          FROM `users`
                          WHERE `id` = $userID";
                $data  = mysql_query($query);
                if(mysql_num_rows($data))
                {
                    $current_user = mysql_fetch_assoc($data);
                    return $current_user;
                }
            }
        }
        return $current_user;
    }