syntax грешка

D

Dobcho

Registered
Здравейте, как мога да оправя тази грешка
Код: 
Parse error: syntax error, unexpected T_VARIABLE in /home/a7518638/public_html/includes/settings.php on line 1
Кодът който използвам изглейда така
Код: 
<?php
$title = "ххх - ххх!";

$host = "mysql16.ххх.com";

$dbuser = "a7518638_race";

$dbpass = "ххх";

$db = "a7518638_race";



$db_conn = mysql_connect("$host", "$dbuser", "$dbpass")

or die("unable to connect to the database");

mysql_select_db("$db", $db_conn) or die("unable to select the database");

mysql_query("SET CHARACTER SET cp1251");



session_start();

$user = $_SESSION['username'];

$selectuser = mysql_query("SELECT * FROM users WHERE username='$user'");

if(mysql_num_rows($selectuser)>0)

{

$urow = mysql_fetch_array($selectuser);

$timenow = time();

$update = mysql_query("UPDATE users SET timeonline='$timenow' WHERE username='$user'");

}

else

{

session_unset();

}



$level11 = '280';

$level12 = '420';

$level13 = '280';

$level14 = '420';

$level15 = '700';

$level16 = '840';

$level17 = '560';

$level18 = '560';

$level19 = '980';



$add1 = '10';

$add2 = '15';

$add3 = '10';

$add4 = '15';

$add5 = '20';

$add6 = '30';

$add7 = '20';

$add8 = '20';

$add9 = '40';



$everylevel = '400';

$everyadd = '12';



if ($HTTP_SESSION_VARS) {extract($HTTP_SESSION_VARS, EXTR_SKIP); }

if ($_SESSION) {extract($_SESSION, EXTR_SKIP); }

if ($HTTP_COOKIE_VARS) {extract($HTTP_COOKIE_VARS, EXTR_SKIP); }

if ($_COOKIE) {extract($_COOKIE, EXTR_SKIP); }

if ($HTTP_POST_VARS) {extract($HTTP_POST_VARS, EXTR_SKIP); }

if ($_POST) {extract($_POST, EXTR_SKIP); }

if ($HTTP_GET_VARS) {extract($HTTP_GET_VARS, EXTR_SKIP); }

if ($_GET) {extract($_GET, EXTR_SKIP); }

if ($HTTP_ENV_VARS) {extract($HTTP_ENV_VARS, EXTR_SKIP); }

if ($_ENV) {extract($_ENV, EXTR_SKIP); } 

?>

Ако е в хост-а (понеже е безплатен..) се извниявам. : )
 
Ами пробвах така, но така не разпознава функцията...

Код: 
Fatal error: Call to undefined function phpsession_start()
 
Код: 
<?php
session_start();
$title = "ххх - ххх!";
$host = "mysql16.ххх.com";
$dbuser = "a7518638_race";
$dbpass = "ххх";
$db = "a7518638_race";



$db_conn = mysql_connect("$host", "$dbuser", "$dbpass")
or die("unable to connect to the database");
mysql_select_db("$db", $db_conn) or die("unable to select the database");
mysql_query("SET CHARACTER SET cp1251");





$user = $_SESSION['username'];
$selectuser = mysql_query("SELECT * FROM users WHERE username='$user'");
if(mysql_num_rows($selectuser)>0)
{
$urow = mysql_fetch_array($selectuser);
$timenow = time();
$update = mysql_query("UPDATE users SET timeonline='$timenow' WHERE username='$user'");
}
else
{
session_unset();
}
$level11 = '280';
$level12 = '420';
$level13 = '280';
$level14 = '420';
$level15 = '700';
$level16 = '840';
$level17 = '560';
$level18 = '560';
$level19 = '980';

$add1 = '10';
$add2 = '15';
$add3 = '10';
$add4 = '15';
$add5 = '20';
$add6 = '30';
$add7 = '20';
$add8 = '20';
$add9 = '40';

$everylevel = '400';
$everyadd = '12';

if ($HTTP_SESSION_VARS) {extract($HTTP_SESSION_VARS, EXTR_SKIP); }
if ($_SESSION) {extract($_SESSION, EXTR_SKIP); }
if ($HTTP_COOKIE_VARS) {extract($HTTP_COOKIE_VARS, EXTR_SKIP); }
if ($_COOKIE) {extract($_COOKIE, EXTR_SKIP); }
if ($HTTP_POST_VARS) {extract($HTTP_POST_VARS, EXTR_SKIP); }
if ($_POST) {extract($_POST, EXTR_SKIP); }
if ($HTTP_GET_VARS) {extract($HTTP_GET_VARS, EXTR_SKIP); }
if ($_GET) {extract($_GET, EXTR_SKIP); }
if ($HTTP_ENV_VARS) {extract($HTTP_ENV_VARS, EXTR_SKIP); }
if ($_ENV) {extract($_ENV, EXTR_SKIP); } 

?>
 
Мерси за отделеното време, но и с този код не се получава отново ми дава
Код: 
Fatal error: Call to undefined function phpsession_start() in /home/a7518638/public_html/includes/settings.php on line 1
тази грешка. Пробвах и на друг хост отново същото е?!
 
PHP: 
<?php
session_start();
ob_start();
$title = "ххх - ххх!";
$host = "mysql16.ххх.com";
$dbuser = "a7518638_race";
$dbpass = "ххх";
$db = "a7518638_race";



$db_conn = mysql_connect("$host", "$dbuser", "$dbpass")
or die("unable to connect to the database");
mysql_select_db("$db", $db_conn) or die("unable to select the database");
mysql_query("SET CHARACTER SET cp1251");





$user = $_SESSION['username'];
$selectuser = mysql_query("SELECT * FROM users WHERE username='$user'");
if(mysql_num_rows($selectuser)>0)
{
$urow = mysql_fetch_array($selectuser);
$timenow = time();
$update = mysql_query("UPDATE users SET timeonline='$timenow' WHERE username='$user'");
}
else
{
session_unset();
}
$level11 = '280';
$level12 = '420';
$level13 = '280';
$level14 = '420';
$level15 = '700';
$level16 = '840';
$level17 = '560';
$level18 = '560';
$level19 = '980';

$add1 = '10';
$add2 = '15';
$add3 = '10';
$add4 = '15';
$add5 = '20';
$add6 = '30';
$add7 = '20';
$add8 = '20';
$add9 = '40';

$everylevel = '400';
$everyadd = '12';

if ($HTTP_SESSION_VARS) {extract($HTTP_SESSION_VARS, EXTR_SKIP); }
if ($_SESSION) {extract($_SESSION, EXTR_SKIP); }
if ($HTTP_COOKIE_VARS) {extract($HTTP_COOKIE_VARS, EXTR_SKIP); }
if ($_COOKIE) {extract($_COOKIE, EXTR_SKIP); }
if ($HTTP_POST_VARS) {extract($HTTP_POST_VARS, EXTR_SKIP); }
if ($_POST) {extract($_POST, EXTR_SKIP); }
if ($HTTP_GET_VARS) {extract($HTTP_GET_VARS, EXTR_SKIP); }
if ($_GET) {extract($_GET, EXTR_SKIP); }
if ($HTTP_ENV_VARS) {extract($HTTP_ENV_VARS, EXTR_SKIP); }
if ($_ENV) {extract($_ENV, EXTR_SKIP); }

?>
 
Обаче и с ob_start(); не стана отново същата грешка ми дава. И нищо немога да измисля.
 
Пробвал ли си с твоя първия код който даде , да закоментираш или изтриеш първия ред? Каква грешка дава тогава?
 
Ехх, значи без session_start(); само с ob_start(); се изчисти от грешки. Мерси на всички които се включиха.
 
Трябва да си влезнал или регистриран, за да отговориш тук.

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