问题
I have a PHP file that is sometimes called from a page that has started a session and sometimes from a page that doesn't have session started. Therefore when I have session_start() on this script I sometimes get the error message for "session already started". For that I've put these lines:
if(!isset($_COOKIE["PHPSESSID"]))
{
session_start();
}
but this time I got this warning message:
Notice: Undefined variable: _SESSION
Is there a better way to check if session has already started?
If I use @session_start will it make things work properly and just shut up the warnings?
回答1:
Recommended way for versions of PHP >= 5.4.0 , PHP 7
if (session_status() == PHP_SESSION_NONE) {
session_start();
}
Reference: http://www.php.net/manual/en/function.session-status.php
For versions of PHP < 5.4.0
if(session_id() == '') {
session_start();
}
回答2:
For versions of PHP prior to PHP 5.4.0:
if(session_id() == '') {
// session isn't started
}
Though, IMHO, you should really think about refactoring your session management code if you don't know whether or not a session is started...
That said, my opinion is subjective, and there are situations (examples of which are described in the comments below) where it may not be possible to know if the session is started.
回答3:
PHP 5.4 introduced session_status(), which is more reliable than relying on session_id().
Consider the following snippet:
session_id('test');
var_export(session_id() != ''); // true, but session is still not started!
var_export(session_status() == PHP_SESSION_ACTIVE); // false
So, to check whether a session is started, the recommended way in PHP 5.4 is now:
session_status() == PHP_SESSION_ACTIVE
回答4:
you can do this, and it's really easy.
if (!isset($_SESSION)) session_start();
Hope it helps :)
回答5:
if (version_compare(phpversion(), '5.4.0', '<')) {
if(session_id() == '') {
session_start();
}
}
else
{
if (session_status() == PHP_SESSION_NONE) {
session_start();
}
}
回答6:
Prior to PHP 5.4 there is no reliable way of knowing other than setting a global flag.
Consider:
var_dump($_SESSION); // null
session_start();
var_dump($_SESSION); // array
session_destroy();
var_dump($_SESSION); // array, but session isn't active.
Or:
session_id(); // returns empty string
session_start();
session_id(); // returns session hash
session_destroy();
session_id(); // returns empty string, ok, but then
session_id('foo'); // tell php the session id to use
session_id(); // returns 'foo', but no session is active.
So, prior to PHP 5.4 you should set a global boolean.
回答7:
For all php version
if ((function_exists('session_status')
&& session_status() !== PHP_SESSION_ACTIVE) || !session_id()) {
session_start();
}
回答8:
Check this :
<?php
/**
* @return bool
*/
function is_session_started()
{
if ( php_sapi_name() !== 'cli' ) {
if ( version_compare(phpversion(), '5.4.0', '>=') ) {
return session_status() === PHP_SESSION_ACTIVE ? TRUE : FALSE;
} else {
return session_id() === '' ? FALSE : TRUE;
}
}
return FALSE;
}
// Example
if ( is_session_started() === FALSE ) session_start();
?>
Source http://php.net
回答9:
Use session_id(), it returns an empty string if not set. It's more reliable than checking the $_COOKIE.
if (strlen(session_id()) < 1) {
session_start();
}
回答10:
if (session_id() === "") { session_start(); }
hope it helps !
回答11:
This should work for all PHP versions. It determines the PHP version, then checks to see if the session is started based on the PHP version. Then if the session is not started it starts it.
function start_session() {
if(version_compare(phpversion(), "5.4.0") != -1){
if (session_status() == PHP_SESSION_NONE) {
session_start();
}
} else {
if(session_id() == '') {
session_start();
}
}
}
回答12:
The only thing you need to do is:
<?php
if(!isset($_SESSION))
{
session_start();
}
?>
回答13:
Not sure about efficiency of such solution, but this is from working project This is also used if you need to define the default language
/**
* Start session
* Fall back to ukrainian language
*/
function valid_session() {
if(session_id()=='') {
session_start();
$_SESSION['lang']='uk';
$_SESSION['lang_id']=3;
}
return true;
}
回答14:
@ before a function call suppresses any errors that may be reported during the function call.
Adding a @ before session_start tells PHP to avoid printing error messages.
For example:
Using session_start() after you've already printed something to the browser results in an error so PHP will display something like "headers cannot be sent: started at (line 12)", @session_start() will still fail in this case, but the error message is not printed on screen.
Before including the files or redirecting to new page use the exit() function, otherwise it will give an error.
This code can be used in all cases:
<?php
if (session_status() !== PHP_SESSION_ACTIVE || session_id() === ""){
session_start();
}
?>
回答15:
On PHP 5.3 this works for me:
if(!strlen(session_id())){
session_name('someSpecialName');
session_start();
}
then you have. If you do not put the not at if statement beginning the session will start any way I do not why.
回答16:
Response BASED on @Meliza Ramos Response(see first response) and http://php.net/manual/en/function.phpversion.php ,
ACTIONS:
- define PHP_VERSION_ID if not exist
- define function to check version based on PHP_VERSION_ID
- define function to openSession() secure
only use openSession()
// PHP_VERSION_ID is available as of PHP 5.2.7, if our
// version is lower than that, then emulate it
if (!defined('PHP_VERSION_ID')) {
$version = explode('.', PHP_VERSION);
define('PHP_VERSION_ID', ($version[0] * 10000 + $version[1] * 100 + $version[2]));
// PHP_VERSION_ID is defined as a number, where the higher the number
// is, the newer a PHP version is used. It's defined as used in the above
// expression:
//
// $version_id = $major_version * 10000 + $minor_version * 100 + $release_version;
//
// Now with PHP_VERSION_ID we can check for features this PHP version
// may have, this doesn't require to use version_compare() everytime
// you check if the current PHP version may not support a feature.
//
// For example, we may here define the PHP_VERSION_* constants thats
// not available in versions prior to 5.2.7
if (PHP_VERSION_ID < 50207) {
define('PHP_MAJOR_VERSION', $version[0]);
define('PHP_MINOR_VERSION', $version[1]);
define('PHP_RELEASE_VERSION', $version[2]);
// and so on, ...
}
}
function phpVersionAtLeast($strVersion = '0.0.0')
{
$version = explode('.', $strVersion);
$questionVer = $version[0] * 10000 + $version[1] * 100 + $version[2];
if(PHP_VERSION_ID >= $questionVer)
return true;
else
return false;
}
function openSession()
{
if(phpVersionAtLeast('5.4.0'))
{
if(session_status()==PHP_SESSION_NONE)
session_start();
}
else // under 5.4.0
{
if(session_id() == '')
session_start();
}
}
回答17:
if (version_compare(PHP_VERSION, "5.4.0") >= 0) {
$sess = session_status();
if ($sess == PHP_SESSION_NONE) {
session_start();
}
} else {
if (!$_SESSION) {
session_start();
}
}
Actually, it is now too late to explain it here anyway as its been solved. This was a .inc file of one of my projects where you configure a menu for a restaurant by selecting a dish and remove/add or change the order. The server I was working at did not had the actual version so I made it more flexible. It's up to the authors wish to use and try it out.
回答18:
Is this code snippet work for you?
if (!count($_SESSION)>0) {
session_start();
}
回答19:
You should reorganize your code so that you call session_start() exactly once per page execution.
回答20:
session_start();
if(!empty($_SESSION['user']))
{
//code;
}
else
{
header("location:index.php");
}
回答21:
PHP_VERSION_ID is available as of PHP 5.2.7, so check this first and if necessary , create it.
session_status is available as of PHP 5.4 , so we have to check this too:
if (!defined('PHP_VERSION_ID')) {
$version = explode('.', PHP_VERSION);
define('PHP_VERSION_ID', ($version[0] * 10000 + $version[1] * 100 + $version[2]));
}else{
$version = PHP_VERSION_ID;
}
if($version < 50400){
if(session_id() == '') {
session_start();
}
}else{
if (session_status() !== PHP_SESSION_ACTIVE) {
session_start();
}
}
回答22:
Based on my practice, before accessing the $_SESSION[] you need to call session_start every time to use the script. See the link below for manual.
http://php.net/manual/en/function.session-start.php
For me at least session_start is confusing as a name. A session_load can be more clear.
回答23:
i ended up with double check of status. php 5.4+
if(session_status() !== PHP_SESSION_ACTIVE){session_start();};
if(session_status() !== PHP_SESSION_ACTIVE){die('session start failed');};
回答24:
You can use the following solution to check if a PHP session has already started:
if(session_id()== '')
{
echo"Session isn't Start";
}
else
{
echo"Session Started";
}
回答25:
This is what I use to determine if a session has started. By using empty and isset as follows:
if (empty($_SESSION) && !isset($_SESSION)) {
session_start();
}
回答26:
Replace session_start(); with:
if (!isset($a)) {
a = False;
if ($a == TRUE) {
session_start();
$a = TRUE;
}
}
来源：https://stackoverflow.com/questions/54533199/mysqli-id-value-can-assign-session-variable-but-already-session-started