Drezard Posted May 24, 2006 Share Posted May 24, 2006 I keep getting a [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]Parse error: syntax error, unexpected $end in C:\Program Files\xampp\htdocs\Website\new_char.php on line 78 [/quote]Everytime i try to run this script. I know that means its a if or a else not with a } on it but i cant work out which one. Please help me.Script:[code]<html><head><title>Untitled Document</title><meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"></head><body><?include('connect.php');$user = $_COOKIE['user']; if (isset($_COOKIE['user'])) { $sql="SELECT * FROM users WHERE user= '$user'"; $result=mysql_query($sql); // Mysql_num_row is counting table rows $count=mysql_num_rows($result); // If result matched $user, table row must be 1 row if($count==1){ //The user is logged in. $sql="SELECT * FROM users WHERE user='$user' and stage='1'"; $result=mysql_query($sql); // Mysql_num_row is counting table rows $count=mysql_num_rows($result); // If result matched $user and 1, table row must be 1 row if($count==1){ //The User hasnt created a character. ?> <form action="<?=$_SERVER['PHP_SELF']?>" method="post"> Enter A Character Name: <input type="text" name="char_name"> <input type="submit" name="submit"> </form> <?php $char_name = empty($_POST['char_name']) ? die ("Please Enter A Character Name") : mysql_escape_string($_POST['char_name']); if(!empty($_POST)){ $char_name = mysql_real_escape_string($_POST['char_name']); $result = mysql_query("SELECT count(*) FROM `char_name` WHERE `char_name` = '$char_name'") or die(mysql_error()); $count = mysql_result($result,0); if($count > 0){ die('This Character Name is already in use. Please select a different one.'); } else { // create query $query = "INSERT INTO users (char_name, char_stage) VALUES ('$char_name', '2')"; // execute query $result = mysql_query($query) or die ("Error in query: $query. ".mysql_error()); // close connection mysql_close($connection); } } else { echo "You have already created a character or have done this part."; }}}else {echo "You must be logged in to view this page.";}if (!isset($_COOKIE['user'])) { echo "You must be logged in to view this page."; } ?></body></html>[/code]- Cheers, Daniel Quote Link to comment https://forums.phpfreaks.com/topic/10323-end-error-php-script-not-working/ Share on other sites More sharing options...
Honoré Posted May 24, 2006 Share Posted May 24, 2006 try adding a second } at line 72[code]echo "You must be logged in to view this page.";}}if (!isset($_COOKIE['user'])) {[/code] Quote Link to comment https://forums.phpfreaks.com/topic/10323-end-error-php-script-not-working/#findComment-38490 Share on other sites More sharing options...
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