session_is_registered()

(PHP 4, PHP 5 < 5.4.0)

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_VARSfor PHP 4.0.6 or less),可以使用isset()检查变量是否在$_SESSION中注册使用。

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

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.

session_register() function is generating warnings. Therefore, instead of using:
<?php
$test = 'Here';
session_register('test');
?>
It is better :
<?php
$_SESSION['test'] = 'Here';
?>

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.

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..

A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}

A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}

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.

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{

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().