PHP 5.6.0beta1 released

session_is_registered

(PHP 4, PHP 5 < 5.4.0)

session_is_registeredÜberprüft, ob eine globale Variable in einer Session registriert ist

Beschreibung

bool session_is_registered ( string $name )

Überprüft, ob eine globale Variable in einer Session registriert ist.

Warnung

Diese Funktion ist seit PHP 5.3.0 DEPRECATED (veraltet) und seit PHP 5.4.0 ENTFERNT.

Parameter-Liste

name

Der Variablenname

Rückgabewerte

Falls in der aktuellen Session eine globale Variable mit dem Namen name registriert ist, gibt session_is_registered() TRUE zurück, andernfalls FALSE.

Anmerkungen

Hinweis:

Benutzen Sie bei Verwendung von $_SESSION (oder $HTTP_SESSION_VARS bei PHP 4.0.6 oder niedriger) isset() um zu prüfen, ob eine Variable in $_SESSION registriert ist.

Achtung

Wenn Sie $_SESSION (oder $HTTP_SESSION_VARS) verwenden, sollten Sie nicht session_register(), session_is_registered() und session_unregister() verwenden.

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User Contributed Notes 7 notes

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1
miguel dot simoes at swirve dot com
11 years ago
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.
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0
someone at the dot inter dot net
1 month ago
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
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0
paimpozhil at gmail dot com
1 year ago
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..
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-1
amol_bhavsar1982 at hotmail dot com
5 years ago
session_register() function is generating warnings. Therefore, instead of using:

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

It is better :

<?php
$_SESSION
['test'] = 'Here';
?>
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-1
someone at the dot inter dot net
1 month ago
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
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-2
Sami
2 months ago
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{
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-5
vectorjohn at example dot com
4 months ago
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.
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