session_is_registered

(PHP 4, PHP 5 < 5.4.0)

session_is_registered检查变量是否在会话中已经注册

说明

session_is_registered ( string $name ) : bool

检查变量是否已经在会话中注册。

Warning

本函数已自 PHP 5.3.0 起废弃并将自 PHP 5.4.0 起移除

参数

name

变量名称。

返回值

session_is_registered() 返回 TRUE 则表示 name 变量已经在当前会话中注册使用,否则返回 FALSE

注释

Note:

如果使用 $_SESSION (或 $HTTP_SESSION_VARS for PHP 4.0.6 or less),可以使用 isset() 检查变量是否在 $_SESSION 中注册使用。

Caution

如果使用 $_SESSION (或 $HTTP_SESSION_VARS), 则不要使用 session_register(), session_is_registered()session_unregister().

User Contributed Notes

pk at majstar dot com 04-Feb-2015 10:08
There's an error in the comment posted by "someone at the dot inter dot net". Correct replacement for function session_is_registered() in PHP 5.4+ is

function session_is_registered($x) {return isset($_SESSION[$x]);}

so just $x instead of '$x' - single quotation mark won't interpolate the variable $x and the function will always return false.
CertaiN 30-Aug-2014 08:15
If your session variables may have NULL value,  use array_key_exists() instead of isset(). If not, use isset() because it performs better than array_key_exists().
someone at the dot inter dot net 06-Mar-2014 11:48
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
someone at the dot inter dot net 06-Mar-2014 11:48
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
Sami 06-Feb-2014 09:56
I can not get the following code to work as it is returning an error on the session_is_registered() and do I have to change anything else in the code

Thank you

if(!session_is_registered('user_name')){

if (isset($_POST['username'])) {
$password1 = clean($_POST["password"]);
$username1 = clean($_POST["username"]);
$password2 = crypt($password1);

$result = @mysql_query ("select * from users where user_name = '".$username1."'");
$lim = @mysql_num_rows( $result );
//|| (strlen($username1) < 6) || (strlen($password1) < 6)
if( ($lim!=0)  ){
$row = @mysql_fetch_array($result);
$password=$row['user_password'];
if (crypt($password1, $password) == $password){
$sql = @mysql_query ("insert into logs (ip, cdate, status) values ('".$REMOTE_ADDR."','". date("Y-m-d H:i:s") ."', 'Login')");
session_register('user_id');
session_register('user_fullname');
session_register('user_name');

$_SESSION['user_name'] = $row['user_name'];
$_SESSION['user_fullname'] = $row['user_fullname'];
$_SESSION['user_id'] = $row['user_id'];
}//if crypt
else{
vectorjohn at example dot com 05-Dec-2013 06:03
The proper equivalent has nothing to do with isset().

Use array_key_exists() because session_is_registered returns true if the variable is in the session at all, even if it's falsy.
paimpozhil at gmail dot com 07-Apr-2013 04:54
For those who have an older application which uses the session_is_registered..and you want to use that in php5.4

You can just define the function if required

function session_is_registered($x)
{
    if (isset($_SESSION['$x']))
    return true;
    else
    return false;
}

May be add the checks to ensure function is not already existing..
amol_bhavsar1982 at hotmail dot com 22-Mar-2009 11:47
session_register() function is generating warnings. Therefore, instead of using:

<?php
$test
= 'Here';
session_register('test');
?>

It is better :

<?php
$_SESSION
['test'] = 'Here';
?>
miguel dot simoes at swirve dot com 13-Jun-2002 04:27
When using PHP 4.2.0 even on the same page where you registered the variable with:

session_register("someVar");

if you try to see if the variable is set and do not assign it a value before, the function used in the previous comment will give the same output.
 This may show that the variable is declared and will not be set until some value is give assign to it.
 I think that this way will give the option to register all the variables used for sure on the process on the first page and using them as the time comes.