PHP Syntax Error... unexpected T_VARIABLE?

Question

Full error:

Parse error: syntax error, unexpected T_VARIABLE in /home/u572186424/public_html/safe.php on line 56

I have been staring at line 56 and cannot figure it out...

     exit();

The whole file follows:

<?php
    include_once("connect.php");
?>

<?
    $sql = "SELECT * FROM users WHERE id='" . mysql_real_escape_string($_SESSION['user_id']) . "'";
    $query = mysql_query($sql) or die(mysql_error());

    $row = mysql_fetch_object($query);
    $id = htmlspecialchars($row->id);
    $userip = htmlspecialchars($row->userip);
    $username = htmlspecialchars($row->username);
    $password = htmlspecialchars($row->password);
    $account_type = htmlspecialchars($row->account_type);
    $money = htmlspecialchars($row->money);
    $exp = htmlspecialchars($row->exp);
    $req_exp = htmlspecialchars($row->req_exp);
    $level = htmlspecialchars($row->level);
    $health = htmlspecialchars($row->health);
    $max_health = htmlspecialchars($row->max_health);
    $lastactive = htmlspecialchars($row->lastactive);
    $energy = htmlspecialchars($row->energy);
    $max_energy = htmlspecialchars($row->max_energy);
    $will = htmlspecialchars($row->will);
    $max_will = htmlspecialchars($row->max_will);
    $brave = htmlspecialchars($row->brave);
    $max_brave = htmlspecialchars($row->max_brave);
    $strength = htmlspecialchars($row->strength);
    $agility = htmlspecialchars($row->agility);
    $guard = htmlspecialchars($row->guard);
    $labor = htmlspecialchars($row->labor);
    $iq = htmlspecialchars($row->iq);
    $rank = htmlspecialchars($row->rank);
?>

<?php
    $sql = "SELECT * FROM sitestats WHERE id='1'";
    $query = mysql_query($sql) or die(mysql_error());
    $row = mysql_fetch_object($query);
    $admins = htmlspecialchars($row->admins);
    $mods = htmlspecialchars($row->mods);
    $hdo = htmlspecialchars($row->hdo);
    $admins_ip = htmlspecialchars($row->admins_ip);
    $mods_ip = htmlspecialchars($row->mods_ip);
    $admin_array = explode("-", $admins);
    $mod_array = explode("-", $mods);
    $hdo_array = explode("-", $hdo);
    $admin_ip_array = explode("-", $admins_ip);
    $mod_ip_array = explode("-", $mods_ip);
?>

<html>
    <body>

    <?
        if(isset($_SESSION['user_id'])) {
            $sql = "UPDATE users SET lastactive=NOW() WHERE id='" . mysql_real_escape_string($_SESSION['user_id']) . "'";
            mysql_query($sql);
        }
        else{
            header("Location: logout.php");
            exit();  // Error here
        }

        $query = "SELECT account_type,rank FROM users WHERE username= "$username";
        $result = mysql_query($query) or die(mysql_error());
        $row = mysql_fetch_array($result);
        if($row['account_type'] == 1){
            $row['rank'] = "Player";
            $rank = "Player";
        }
        elseif($row['account_type'] == 2){
            $row['rank'] = "VIP";
            $rank = "VIP";
        }
        elseif($row['account_type'] == 3){
            $row['rank'] = "HDO";
            $rank = "HDO";
        }
        elseif($row['account_type'] == 4){
            $row['rank'] = "Moderator";
            $rank = "Moderator";
        }
        elseif($row['account_type'] == 5){
            $row['rank'] = "Admin";
            $rank = "Admin";
        }
        elseif($row['account_type'] == 6){
            $row['rank'] = "Owner";
            $rank = "Owner";
        }
    ?>

</body>
</html>

How can I fix this problem?

yo, indent that a little so we can actually read it? Also, reduce the code until the error goes away. When it does, what you just removed caused the error. Typically you do this first because you tend to find the problem before needing to post a question on it — Mike 'Pomax' Kamermans
– Mike 'Pomax' Kamermans, Commented Nov 14, 2013 at 18:17

AbraCadaver · Accepted Answer · 2013-11-14 18:17:35Z

2

The syntax highlighting shows you. Problem is the extra quote " here:

$query = "SELECT account_type,rank FROM users WHERE username= "$username";

Try:

$query = "SELECT account_type,rank FROM users WHERE username= '$username'";

answered Nov 14, 2013 at 18:17

AbraCadaver

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4 Comments

Edson Horacio Junior Over a year ago

if your string is using double quotes, you can also write a variable without having to close the quotes, concating and opening again, like this: $query = "SELECT something FROM someTable WHERE condition = {$condition}";

AbraCadaver Over a year ago

And notice the use of the single quotes ' as I assume username is text/varchar in the db.

Jacob Over a year ago

Yes, thanks for all the feedback. I can't believe I didn't see that :P

AbraCadaver Over a year ago

It was easy to spot the color change in the code that you posted, so try and get an editor with good syntax highlighting.

regulus · Accepted Answer · 2013-11-14 18:17:58Z

0

There are no dots when you concatenate the query and $username:

$query = "SELECT account_type,rank FROM users WHERE username= "$username";

answered Nov 14, 2013 at 18:17

regulus

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2 Comments

Alex K Over a year ago

You don't need to concatenate a string with a variable containing a string in PHP. You can just do: $query = "SELECT account_type,rank FROM users WHERE username=$username";

regulus Over a year ago

That's right, you either have to put your variable inside quotation marks, or concatenate it. Not the half way.

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