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(PHP 4, PHP 5 < 5.4.0)
session_is_registered — 检查变量是否在会话中已经注册
$name
)检查变量是否已经在会话中注册。
本函数已自 PHP 5.3.0 起废弃并将自 PHP 5.4.0 起移除。
name
变量名称。
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
。
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。
如果使用 $_SESSION (或 $HTTP_SESSION_VARS ), 则不要使用 session_register() , session_is_registered() 和 session_unregister() .
[#1] pk at majstar dot com [2015-02-04 22:08:49]
There's an error in the comment posted by "someone at the dot inter dot net". Correct replacement for function session_is_registered() in PHP 5.4+ is
function session_is_registered($x) {return isset($_SESSION[$x]);}
so just $x instead of '$x' - single quotation mark won't interpolate the variable $x and the function will always return false.
[#2] CertaiN [2014-08-30 08:15:40]
If your session variables may have NULL value, use array_key_exists() instead of isset(). If not, use isset() because it performs better than array_key_exists().
[#3] someone at the dot inter dot net [2014-03-06 11:48:30]
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
[#4] someone at the dot inter dot net [2014-03-06 11:48:28]
A simple one-line function to emulate this in later versions of PHP:
function session_is_registered($x){return isset($_SESSION['$x']);}
[#5] Sami [2014-02-06 09:56:54]
I can not get the following code to work as it is returning an error on the session_is_registered() and do I have to change anything else in the code
Thank you
if(!session_is_registered('user_name')){
if (isset($_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'] = $row['user_id'];
}//if crypt
else{
[#6] vectorjohn at example dot com [2013-12-05 18:03:50]
The proper equivalent has nothing to do with isset().
Use array_key_exists() because session_is_registered returns true if the variable is in the session at all, even if it's falsy.
[#7] paimpozhil at gmail dot com [2013-04-07 16:54:15]
For those who have an older application which uses the session_is_registered..and you want to use that in php5.4
You can just define the function if required
function session_is_registered($x)
{
if (isset($_SESSION['$x']))
return true;
else
return false;
}
May be add the checks to ensure function is not already existing..
[#8] amol_bhavsar1982 at hotmail dot com [2009-03-22 23:47:56]
session_register() function is generating warnings. Therefore, instead of using:
<?php
$test = 'Here';
session_register('test');
?>
It is better :
<?php
$_SESSION['test'] = 'Here';
?>
[#9] miguel dot simoes at swirve dot com [2002-06-13 04:27:04]
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.