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_is_registered
(PHP 4, PHP 5 < 5.4.0)
session_is_registered — 変数がセッションに登録されているかどうかを調べる
説明
session_is_registered
( string
$name
) : boolグローバル変数がセッションに登録されているかどうかを調べます。
警告
この関数は PHP 5.3.0 で 非推奨となり、 PHP 5.4.0 で削除されました。
パラメータ
-
name -
変数名。
返り値
session_is_registered() は、
name
という名前のグローバル変数が現在のセッションに登録されている場合に
TRUE、それ以外の場合に FALSE を返します。
注意
注意:
$_SESSION が使用されている場合、ある変数が $_SESSION に登録されているかを確認するために isset() を使用してください。
警告
$_SESSION (もしくは $HTTP_SESSION_VARS) を使用している場合、 session_register(), session_is_registered(), session_unregister() を使用しないでください。
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User Contributed Notes 9 notes
pk at majstar dot com ¶
9 years ago
amol_bhavsar1982 at hotmail dot com ¶
15 years ago
session_register() function is generating warnings. Therefore, instead of using:
<?php
$test = 'Here';
session_register('test');
?>
It is better :
<?php
$_SESSION['test'] = 'Here';
?>
someone at the dot inter dot net ¶
10 years ago
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
paimpozhil at gmail dot com ¶
11 years 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['$x']))
return true;
else
return false;
}
May be add the checks to ensure function is not already existing..
someone at the dot inter dot net ¶
10 years ago
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
miguel dot simoes at swirve dot com ¶
22 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.
vectorjohn at example dot com ¶
11 years 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.
Sami ¶
10 years 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{
