PHP MySQL: Parse error: syntax error, unexpected T_VARIABLE

Question

I'm creating a mysql database for my android app to save and load a top 10 highscore system. I've some knowledge of php and mysql over the years however I'm no expert. Any help is appreciated:

The full code is:

<?php
define("DB_DSN",'yourdbname');
define("DB_HOST",'localhost');
define("DB_USER",'yourdblogin');
define("DB_PASS",'yourdbpass');
 
// Connecting, selecting database
$link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die('Could not connect: ' . mysql_error());
mysql_select_db(DB_DSN) or die('Could not select database');
 
if(isset($_GET)) {
    $playername = base64_decode($_GET["playername"]);
    $password = base64_decode($_GET["password"]);
    $query = 'SELECT * FROM players WHERE playername="' . mysql_real_escape_string($playername) . '" or email="' . mysql_real_escape_string($loginid) . '"';
    $dbresult = mysql_query($query, $link);
    if (!$dbresult) {
        //echo "query failed";
        $result = array();
        $result["result"] = 403;
        $result["message"] = mysql_error();
        echo json_encode($result);
        mysql_free_result($dbresult);
        exit;
    }
    $player = mysql_fetch_array($dbresult, MYSQL_ASSOC);
    if (strcmp($player["password"],md5($password)) == 0) {
        $result = array();
        $result["result"] = 200;
        $result["message"] = "Success";
        $result["playername"] = $player["playername"];
        $result["firstname"] = $player["firstname"];
        $result["lastname"] = $player["lastname"]; 
        $query = sprintf("UPDATE players SET lastlogin=NOW() WHERE id=%s;", $player["id"]);
        $uresult = mysql_query($query, $link);
        if ($uresult) {
            //code if your update failed.  Doesn't really impact what we are doing. so do nothing.
        }
        echo json_encode($result);
    } else {
        //echo "password mismatch";
        $result = array();
        $result["result"] = 403;
        $result["message"] = "Forbidden";
        echo json_encode($result);
    }
} else {
    $result = array();
    $result["result"] = 400;
    $result["message"] = "Bad Request";
    echo json_encode($result);
}
exit;

and it breaks on line 8:

$link = mysql_connect(DB_HOST, DB_USER, DB_PASS)
    or die('Could not connect: ' . mysql_error());

Answer 1

Not sure if your fixed this or not. It was a long time ago.

but this is how you could easily find errors in either your php code or your query.

1. To start add the following to the top of your php file (right after opening [

   error_reporting(E_ALL); ini_set('display_errors', 'on');

This will show you what php errors you have on the page.

2. Then check your query error by using

   $sql = your sql query
   $query = mysqli_query($link, "$sql") or die(mysqli_error($link);
   

The second will print the mysql error, which in turn you can then use to fix your problem .

You can always use the above to find problems.