Check if PHP session has already started

Question

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?

Answer 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();
}

Answer 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.

Answer 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

Answer 4

you can do this, and it's really easy.

if (!isset($_SESSION)) session_start();

Answer 5

if (version_compare(phpversion(), '5.4.0', '<')) {
     if(session_id() == '') {
        session_start();
     }
 }
 else
 {
    if (session_status() == PHP_SESSION_NONE) {
        session_start();
    }
 }

Answer 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.

Answer 7

For all php version

if ((function_exists('session_status') 
  && session_status() !== PHP_SESSION_ACTIVE) || !session_id()) {
  session_start();
}

Answer 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

Answer 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();
}

Answer 10

if (session_id() === "") { session_start(); }

hope it helps !

Answer 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();
    }
  }
}

Answer 12

The only thing you need to do is:

<?php
if(!isset($_SESSION))
{
session_start();
}
?>

Answer 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;
  }

Answer 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(); 
        }
    ?>

Answer 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.

Answer 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();
        }
    }

Answer 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.

Answer 18

Is this code snippet work for you?

if (!count($_SESSION)>0) {
    session_start();
}

Answer 19

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();
}

Answer 20

session_start();
if(!empty($_SESSION['user']))
{     
  //code;
}
else
{
    header("location:index.php");
}

Answer 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();
    }
}

Answer 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.

Answer 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');};

Answer 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";
}

Answer 25

You should reorganize your code so that you call session_start() exactly once per page execution.

Answer 26

Replace session_start(); with:

if (!isset($a)) {
    a = False;
    if ($a == TRUE) {
        session_start();
        $a = TRUE;
    }
}