Parse error: syntax error, unexpected ';'/ XXX on line XX

Discussion in 'PHP' started by saherboy, Jun 2, 2012.

  1. #1
    Hi every one
    i have the following error
    Parse error: syntax error, unexpected ';' hosting directory complete.php on line 50
    the code is
    <?
    include('../../config.php');
    require_once "../../includes/pluggable.php";
    foreach( glob("../../plugins/*/index.php")  as $plugin) {
      require_once($plugin); 
    }  
    foreach($_POST as $key => $value) {
    	$posts[$key] = filter($value);
    }
    if(isset($posts['me'])){
    if(isset($_GET['step']) && $_GET['step'] == "skip"){
    mysql_query("INSERT INTO `followed` (user_id, site_id) VALUES('{$posts['me']}', '{$posts['him']}')");
    echo "Done!";
    }else{
    $dbres1	= mysql_query("SELECT * FROM `users` WHERE `id`='{$posts['me']}'");
    $data3 = mysql_fetch_object($dbres1);
    $site2 = mysql_fetch_object(mysql_query("SELECT * FROM `twitter` WHERE `id`='{$posts['him']}'"));
    $id = explode("/", $site2->url);
    $id = $id[count($id) - 1];
    $url  = "http://api.twitter.com/1/friendships/exists.json?screen_name_a=" . $data3->twitter . "&screen_name_b=".$id;
    $page = get_data($url);
    $json_a = json_decode($page,true);
    
    
    $plused1 = mysql_query("SELECT * FROM `followed` WHERE `site_id`='{$posts['him']}' AND `user_id`='{$posts['me']}'");
    $plused  = mysql_num_rows($plused1);
    if ($page == FALSE) { 
        echo "Error: Wir k?nnen keine Follower best?tigen!";
    }else if($plused > 0){
    echo "Error: Sie haben bereits gefollowed @".$id."!";
    }else if (strpos($page, "true") !== FALSE) {
    $coins = number_format($site2->cpc - 1);
    $bonuscoins = hook_filter('bonus_coins',$coins);
    
    mysql_query("UPDATE `users` SET `coins`=`coins`+'{$bonuscoins}' WHERE `id`='{$posts['me']}'");
    mysql_query("UPDATE `twitter` SET `exchanges`=`exchanges`+'1' WHERE `id`='{$posts['him']}'");
    mysql_query("UPDATE `users` set `coins`=`coins`-'{$site2->cpc}' WHERE `id`='{$site2->user}'");
    mysql_query("INSERT INTO `followed` (user_id, site_id) VALUES('{$posts['me']}', '{$posts['him']}')");
    
    $aff1 = mysql_query("SELECT ref FROM `users` WHERE `id`='{$posts['me']}'");
    $aff = mysql_fetch_object($aff1);
    if($aff->ref > 0){
    $coins = $coins * $site->refbonus;
    mysql_query("UPDATE `users` SET `coins`=`coins`+'$coins' WHERE `id`='{$aff->ref}'");
    }
    
    echo "Sie haben gefollowed @".$id."! Sie erhalten {$coins} Coins!";
    }else{
    echo "Twitter meldet: Sie haben noch nicht gefollowed @".$id."! Versuchen sie es erneut wenn sie sich eingeloggt haben" . $data3->twitter . 
    ?>
    
    }
    }
    ?>
    
    Code (markup):
    i hope you can help me to fix this error if possible
     
    Solved! View solution.
    saherboy, Jun 2, 2012 IP
  2. akhileshbc

    akhileshbc Active Member

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    #2
    Check the last "echo" statement. You are missing a semicolon at the end (instead of that dot operator). :)
     
    akhileshbc, Jun 2, 2012 IP
  3. saherboy

    saherboy Well-Known Member

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    #3
    Hi
    just do it as follows
    <?
    include('../../config.php');
    require_once "../../includes/pluggable.php";
    foreach( glob("../../plugins/*/index.php")  as $plugin) {
      require_once($plugin); 
    }  
    foreach($_POST as $key => $value) {
    	$posts[$key] = filter($value);
    }
    if(isset($posts['me'])){
    if(isset($_GET['step']) && $_GET['step'] == "skip"){
    mysql_query("INSERT INTO `followed` (user_id, site_id) VALUES('{$posts['me']}', '{$posts['him']}')");
    echo "Done!";
    }else{
    $dbres1	= mysql_query("SELECT * FROM `users` WHERE `id`='{$posts['me']}'");
    $data3 = mysql_fetch_object($dbres1);
    $site2 = mysql_fetch_object(mysql_query("SELECT * FROM `twitter` WHERE `id`='{$posts['him']}'"));
    $id = explode("/", $site2->url);
    $id = $id[count($id) - 1];
    $url  = "http://api.twitter.com/1/friendships/exists.json?screen_name_a=" . $data3->twitter . "&screen_name_b=".$id;
    $page = get_data($url);
    $json_a = json_decode($page,true);
    
    
    $plused1 = mysql_query("SELECT * FROM `followed` WHERE `site_id`='{$posts['him']}' AND `user_id`='{$posts['me']}'");
    $plused  = mysql_num_rows($plused1);
    if ($page == FALSE) { 
        echo "Error: Wir k?nnen keine Follower best?tigen!";
    }else if($plused > 0){
    echo "Error: Sie haben bereits gefollowed @".$id."!";
    }else if (strpos($page, "true") !== FALSE) {
    $coins = number_format($site2->cpc - 1);
    $bonuscoins = hook_filter('bonus_coins',$coins);
    
    mysql_query("UPDATE `users` SET `coins`=`coins`+'{$bonuscoins}' WHERE `id`='{$posts['me']}'");
    mysql_query("UPDATE `twitter` SET `exchanges`=`exchanges`+'1' WHERE `id`='{$posts['him']}'");
    mysql_query("UPDATE `users` set `coins`=`coins`-'{$site2->cpc}' WHERE `id`='{$site2->user}'");
    mysql_query("INSERT INTO `followed` (user_id, site_id) VALUES('{$posts['me']}', '{$posts['him']}')");
    
    $aff1 = mysql_query("SELECT ref FROM `users` WHERE `id`='{$posts['me']}'");
    $aff = mysql_fetch_object($aff1);
    if($aff->ref > 0){
    $coins = $coins * $site->refbonus;
    mysql_query("UPDATE `users` SET `coins`=`coins`+'$coins' WHERE `id`='{$aff->ref}'");
    }
    
    echo "Sie haben gefollowed @".$id."! Sie erhalten {$coins} Coins!";
    }else{
    echo "Twitter meldet: Sie haben noch nicht gefollowed @".$id."! Versuchen sie es erneut wenn sie sich eingeloggt haben" . $data3->twitter ;
    ?>
    
    }
    }
    ?>
    Code (markup):
    edit this line
    echo "Twitter meldet: Sie haben noch nicht gefollowed @".$id."! Versuchen sie es erneut wenn sie sich eingeloggt haben" . $data3->twitter ;
    Code (markup):
    and got new error
    Parse error: syntax error, unexpected $end in web hosting files complete.php on line 55
     
    saherboy, Jun 2, 2012 IP
  4. akhileshbc

    akhileshbc Active Member

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    #4
    Try removing the second last closing tag of php code ("?>"). The one just beneath the echo line which we have edited.

    And also, try changing the very first line from this:
    <?
    PHP:
    to this:
    <?php
    PHP:
    Hope it would work. :)
     
    akhileshbc, Jun 2, 2012 IP
  5. saherboy

    saherboy Well-Known Member

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    #5
    thanks for your interesting
    it is not working :(
    Same error as before $end
     
    saherboy, Jun 2, 2012 IP
  6. #6
    If you format your code properly with indents you can more easily see any problems you have. Try the code below.

    <?
    include('../../config.php');
    require_once("../../includes/pluggable.php");
    
    foreach( glob("../../plugins/*/index.php")  as $plugin) {
    	require_once($plugin); 
    } 
     
    foreach($_POST as $key => $value) {
    	$posts[$key] = filter($value);
    }
    if(isset($posts['me'])){
    	if(isset($_GET['step']) && $_GET['step'] == "skip"){
    		mysql_query("INSERT INTO `followed` (user_id, site_id) VALUES('{$posts['me']}', '{$posts['him']}')");
    		echo "Done!";
    	}else{
    		$dbres1	= mysql_query("SELECT * FROM `users` WHERE `id`='{$posts['me']}'");
    		$data3 = mysql_fetch_object($dbres1);
    		$site2 = mysql_fetch_object(mysql_query("SELECT * FROM `twitter` WHERE `id`='{$posts['him']}'"));
    		$id = explode("/", $site2->url);
    		$id = $id[count($id) - 1];
    		$url  = "http://api.twitter.com/1/friendships/exists.json?screen_name_a=" . $data3->twitter . "&screen_name_b=".$id;
    		$page = get_data($url);
    		$json_a = json_decode($page,true);
    
    		$plused1 = mysql_query("SELECT * FROM `followed` WHERE `site_id`='{$posts['him']}' AND `user_id`='{$posts['me']}'");
    		$plused  = mysql_num_rows($plused1);
    		if ($page == FALSE) { 
    			echo "Error: Wir k?nnen keine Follower best?tigen!";
    		}else if($plused > 0){
    			echo "Error: Sie haben bereits gefollowed @".$id."!";
    		}else if (strpos($page, "true") !== FALSE) {
    			$coins = number_format($site2->cpc - 1);
    			$bonuscoins = hook_filter('bonus_coins',$coins);
    
    			mysql_query("UPDATE `users` SET `coins`=`coins`+'{$bonuscoins}' WHERE `id`='{$posts['me']}'");
    			mysql_query("UPDATE `twitter` SET `exchanges`=`exchanges`+'1' WHERE `id`='{$posts['him']}'");
    			mysql_query("UPDATE `users` set `coins`=`coins`-'{$site2->cpc}' WHERE `id`='{$site2->user}'");
    			mysql_query("INSERT INTO `followed` (user_id, site_id) VALUES('{$posts['me']}', '{$posts['him']}')");
    
    			$aff1 = mysql_query("SELECT ref FROM `users` WHERE `id`='{$posts['me']}'");
    			$aff = mysql_fetch_object($aff1);
    			if($aff->ref > 0){
    				$coins = $coins * $site->refbonus;
    				mysql_query("UPDATE `users` SET `coins`=`coins`+'$coins' WHERE `id`='{$aff->ref}'");
    			}	
    			echo "Sie haben gefollowed @".$id."! Sie erhalten {$coins} Coins!";
    		}else{
    			echo "Twitter meldet: Sie haben noch nicht gefollowed @".$id."! Versuchen sie es erneut wenn sie sich eingeloggt haben" . $data3->twitter ;
    		}
    	}
    }
    ?>
    PHP:
     
    PK-Host, Jun 2, 2012 IP
  7. saherboy

    saherboy Well-Known Member

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    #7
    Thanks it work
    :)
    a lot of wishes for you
     
    saherboy, Jun 2, 2012 IP