what does Parse error: syntax error, unexpected T_VARIABLE acutally mean

Question

hello I am getting this error

Parse error: syntax error, unexpected T_VARIABLE in F:\xampp\htdocs\jim\home.php on line 6

$ses_id = session_id();
$user =$_SESSION['user'] 
$sesssion_id=$_SESSION['sessionid'] 
if($user == "" || $sesssion_id != $ses_id)
{
    echo "go back";
}

can anyone tell me what exactly this error mean ? and please see where i am doing wrong

add two `;` semi-colons in this two statements ...`$user =$_SESSION['user'] $sesssion_id=$_SESSION['sessionid']` — Tun Zarni Kyaw, Mar 16 '14 at 20:04
if you are just looking what `T_VARIABLE` means ... view this one http://stackoverflow.com/questions/7640419/why-the-t-in-phps-unexpected-t-variable — Tun Zarni Kyaw, Mar 16 '14 at 20:09

Gauthier · Answer 1 · 2016-11-29T16:17:08.487

Unexpected T_VARIABLE usally means that php was parsing your code, but something unexpected came up as the next characters.

So when parsing your code

$user = $_SESSION['user'] $sesssion_id = $_SESSION['sessionid']

The php parser will get past "$user = $_SESSION['user']", but it expects the assignment to finish with another semi-colon. Without the semi-color it is expecting another assigntment like concatenation or math, however it runs into another assignment and will throw the Unexpected T_VARIABLE.

$sesssion_id = $_SESSION['sessionid'] This is your next block of code, and as everyone else has suggested, the way of fixing the parsing error (the unexcepted T_VARIABLE) is to add a semicolon ( ; ) to let the parser know to start parsing another statement.

score 2 · Answer 2 · answered Mar 16 '14 at 20:03

2

The following is wrong

$ses_id = session_id(); $user =$_SESSION['user'] $sesssion_id=$_SESSION['sessionid']

You need to add ; to end the statement as

$ses_id = session_id(); 
$user =$_SESSION['user'];
 $sesssion_id=$_SESSION['sessionid'];

answered Mar 16 '14 at 20:03

Abhik Chakraborty

44,654
6
52
63

score 0 · Answer 3 · answered Mar 16 '14 at 20:03

0

You've missed ;

  $ses_id = session_id(); 
  $user =$_SESSION['user'] ;
  $sesssion_id=$_SESSION['sessionid'];

  if($user == "" || $sesssion_id != $ses_id){
     echo "go back";
  }

answered Mar 16 '14 at 20:03

falinsky

7,229
3
32
56

what does Parse error: syntax error, unexpected T_VARIABLE acutally mean

3 Answers3