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PHP - Unexpected $end
Similar TutorialsHi, could someone please tell me why I'm getting an unexpected { error in the following code. I have went over it several times and everything seems to be matching. Code: [Select] <?php require_once("functions.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <?php DatabaseConnection(); $query= "SELECT * FROM treats"; $result_set= mysql_query($query); /*if( $result_set = mysql_query($query) ) { while( $products = mysql_fetch_row($result_set) ) { echo $products[0]; echo $products[1]; echo $products[2]; echo $products[3]; echo $products[4]; echo "<img src=\"{$products[5]}\">"; // assuming this is where the image url is } } //print_r(mysql_fetch_row($result_set));*/ $output = "<table>"; while($row = mysql_fetch_array($result)) { $productDetail1['product_id']; $productDetail2['product_title']; $productDetail3['product_Description']; $productDetail4['price']; $productDetail5['product_pic']; $output .= (" <tr> <td>".$productDetail1['product_id'] ."</td> </tr> ") } $output .= "</table>"; ?> </body> </html> Am I blind? I don't see the problem with this code. It says: Parse error: syntax error, unexpected '}' in /data/21/2/40/160/2040975/user/2235577/htdocs/edit_user1.php on line 35 Code: [Select] <html> <body> <?php if ( (isset($_GET['id'])) && (is_numeric($_GET['id'])) ) { $id = $_GET['id']; } elseif ( (isset($_POST['id'])) && (is_numeric($_POST['id'])) ) { $id = $_POST['id']; } else { echo 'you have reached this page in error. no variable passed'; exit(); } // that was the part to check for passed variables. This is what does something with it require ('databaseconnect.php'); if (isset($_POST['submitted'])) { $errors = array(); } if (empty($_POST['scientific name'])) { $errors[] = 'you did not enter a scientific name' } else { $sn = escape_data($_POST['scientific_name']); } // now the common name if (empty($_POST['common_name_english'])) { $errors[] = 'you did not enter a common name' } else { $cn = escape_data($_POST['scientific_name_english']); } // now make changes if (empty($errors)) { $query = "UPDATE table SET plant_name='$id', scientific_name='$sn', common_name_english='$cn' WHERE plant_name=$id"; $result = $mysql_query ($query); if (mysql_affected_row() == 1) { echo 'edit a plant<br> The plant has edited' } else { echo 'system error<br> the plant could not be edited due to a system error.'; echo mysql_error() . 'query:' . $query ; exit(); } } else { echo 'error<br> something already insystem or did not work.'; } } else { echo 'error<br>'; foreach ($errors as $msg) { echo " - $msg<br>"; } echo ' please try again'; } // end of if } // endo of submit condition // always show form $query = "SELECT scientific_name, Common_name_english FROM table WHERE plant_name=$id"; $result = $mysql_query ($query); if(mysql_num_rows($result) == 1) { $row mysql fetch_array ($result); echo 'edit a user' <form action="edit_user1.php" method="post"> scientific name: <input type="text" name="scientific_name" size="45" value="' . $row[scientific_name] . '"><br> common name: <input type="text" name="common_name_english" size="45" value="' . $row[common_name_english] . '"><br> <input type="submit" name="submit" value="submit" /><br> <input type="hidden" name="submitted" value="TRUE" /> <input type="hidden" name="id" value="' . $id . '"/> </form>'; } else { echo 'page error'; } mysql_close(); </body> </html> I know this is probably a missing curly bracket or some other syntax, but I can't seem to spot it, anyone see it? i get this error btw. Parse error: syntax error, unexpected $end in C:\Program Files\xampp\htdocs\cameo\login.php on line 39 <?php session_start(); //sql variables $server = "localhost"; $user = "root"; $pass = ""; $db = "cameo"; //iff user isn't logged in, try to log them in if(!isset($_SESSION['username'])){ if(isset($_POST['submit'])){ //connect to database $dbc = mysqli_connect($server, $user, $pass, $db); //grab entered form data $user_username = mysqli_real_escape_string($dbc, trim($_POST['username'])); $user_password = mysqli_real_escape_string($dbc, trim($_POST['username'])); //if both username and password are entered then find a match in the database if((!empty($user_username)) && (!empty($user_password))) { //look up the username and password in the database $query = "SELECT username FROM cameo WHERE username = '$user_username' AND '$user_password'"; $data = mysqli_query($dbc, $query); //authenticate if data matches a row if(mysqli_num_rows($data) == 1){ //login is okay $row = mysqli_fetch_array($data); $_SESSION['username'] = $row['username']; $_SESSION['is_admin'] = $row['is_admin']; header('Location:index.php'); } else { //username and password are incorrect or missing, so send a message $error_msg = 'Sorry, you must enter a valid username and password to log in.'; } } } ?> Hi, could someone please tell me why I'm getting unexpected $end error in the following code? <?php function documentType(){ echo <<<HEREDOC <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" ` ` "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> } HEREDOC; ?> Just trying to do a basic query... not working and don't know why: here is inserts.php: <?php $username="wormste1_barry"; $password="barry"; $database="wormste1_barry"; $CarName=$_POST['CarName']; $CarTitle=$_POST['CarTitle']; $CarPrice=$_POST['CarPrice']; $CarMiles=$_POST['CarMiles']; $CarDescription=$_POST['CarDescription']; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO tablename VALUES ('','$CarName','$CarTitle','$CarPrice','$CarMiles','$CarDescription'); mysql_query($query); mysql_close(); ?> That parses the simple form of : form.html: <HTML> <HEAD> </HEAD> <BODY> <form action="inserts.php" method="post"> car Name: <input type="text" name="CarName"><br> Car Title: <input type="text" name="CarTitle"><br> Car Price: <input type="text" name="CarPrice"><br> Car Miles: <input type="text" name="CarMiles"><br> Car Description: <input type="text" name="CarDescription"><br> <input type="Submit"> </form> </BODY> </HTML> I get the error: Parse error: syntax error, unexpected $end in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 23 Don't know whats wrong? hello i keep getting an error Parse error: syntax error, unexpected $end in /home/cookbook/public_html/cookbook.php on line 144 can someplease explain it to me? here is the code [list type=decimal] [li][/li] [li][/li] [/list]<?php //include("include/session.php"); @session_start(); // Start_session, check if user is logged in or not, and connect to the database all in one included file include_once("scripts/checkuserlog.php"); // Include the class files for auto making links out of full URLs and for Time Ago date formatting include_once("wi_class_files/autoMakeLinks.php"); include_once ("wi_class_files/agoTimeFormat.php"); // Create the two objects before we can use them below in this script $activeLinkObject = new autoActiveLink; $myObject = new convertToAgo; ?> <?php // Include this script for random member display on home page include_once "scripts/homePage_randomMembers.php"; ?> <?php $sql_blabs = mysql_query("SELECT id, mem_id, the_blab, blab_date FROM blabbing ORDER BY blab_date DESC LIMIT 30"); $blabberDisplayList = ""; // Initialize the variable here while($row = mysql_fetch_array($sql_blabs)){ $blabid = $row["id"]; $uid = $row["mem_id"]; $the_blab = $row["the_blab"]; $notokinarray = array("fag", "gay", "shit", "fuck", "stupid", "idiot", "asshole", "cunt", "douche"); $okinarray = array("sorcerer", "grey", "shug", "farg", "smart", "awesome guy", "asshole", "cake", "dude"); $the_blab = str_replace($notokinarray, $okinarray, $the_blab); $the_blab = ($activeLinkObject -> makeActiveLink($the_blab)); $blab_date = $row["blab_date"]; $convertedTime = ($myObject -> convert_datetime($blab_date)); $whenBlab = ($myObject -> makeAgo($convertedTime)); //$blab_date = strftime("%b %d, %Y %I:%M:%S %p", strtotime($blab_date)); // Inner sql query $sql_mem_data = mysql_query("SELECT id, username, firstname, lastname FROM myMembers WHERE id='$uid' LIMIT 1"); //die($sql_mem_data); while($row = mysql_fetch_array($sql_mem_data)){ $uid = $row["id"]; $username = $row["username"]; $firstname = $row["firstname"]; if ($firstname != "") {$username = $firstname; } // (I added usernames late in my system, this line is not needed for you) /////// Mechanism to Display Pic. See if they have uploaded a pic or not ////////////////////////// $ucheck_pic = "members/$uid/image01.jpg"; $udefault_pic = "members/0/image01.jpg"; if (file_exists($ucheck_pic)) { $blabber_pic = '<div style="overflow:hidden; width:40px; height:40px;"><img src="' . $ucheck_pic . '" width="40px" border="0" /></div>'; // forces picture to be 100px wide and no more } else { $blabber_pic = "<img src=\"$udefault_pic\" width=\"40px\" height=\"40px\" border=\"0\" />"; // forces default picture to be 100px wide and no more } $blabberDisplayList .= ' <table width="100%" align="center" cellpadding="4" bgcolor="#CCCCCC"> <tr> <td width="7%" bgcolor="#FFFFFF" valign="top"><a href="profile.php?id=' . $uid . '">' . $blabber_pic . '</a> </td> <td width="93%" bgcolor="#EFEFEF" style="line-height:1.5em;" valign="top"><span class="greenColor textsize10">' . $whenBlab . ' <a href="profile.php?id=' . $uid . '">' . $username . '</a> said: </span><br /> ' . $the_blab . '</td> </tr> </table>'; } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1" /> <meta name="Description" content="Web Intersect is a deft combination of powerful free open source software for social networking, mixed with insider guidance and tutorials as to how it is made at its core for maximum adaptability. The goal is to give you a free website system that has a network or community integrated into it to allow people to join and interact with your website when you have the need." /> <meta name="Keywords" content="web intersect, how to build community, build social network, how to build website, learn free online, php and mysql, internet crossroads, directory, friend, business, update, profile, connect, all, website, blog, social network, connecting people, youtube, myspace, facebook, twitter, dynamic, portal, community, technical, expert, professional, personal, find, school, build, join, combine, marketing, optimization, spider, search, engine, seo, script" /> <title>CookBookers</title> <link href="style/main.css" rel="stylesheet" type="text/css" /> <link rel="icon" href="favicon.ico" type="image/x-icon" /> <link rel="shortcut icon" href="favicon.ico" type="image/x-icon" /> <script src="js/jquery-1.4.2.js" type="text/javascript"></script> <style type="text/css"> #Layer1 { height:210px; background-image: url(images/top_container_bg_recipes_new.gif); } body { background-color: #3c60a4; } </style> </head> <body> <?php include_once "header_template.php"; ?> <center> <table cellpadding="0px" cellspacing="0px" style="border:0px solid #666666;" width="950"> <tr> <td> <table width="95%" height="22" border="0" align="center" cellpadding="10" cellspacing="0" style="background-color:#F2F2F2; border:0px solid #666666;"> <tr> <td style="padding-left:45px;"> <?php //die($_SESSION['username']); $qryUsers = "SELECT * from recipies WHERE user='".$_SESSION['username']."'"; //die($qryUsers); $rsUsers = @mysql_query($qryUsers) or die(mysql_error()); ?> <table width="100%" border="0" cellspacing="0" cellpadding="0"> <tr> <td width="16%"><h3>All Recipies of </h3></td> <td width="84%"><h3><?php print $_SESSION['username']; ?></h3></td> </tr> <tr> <td> </td> <td> </td> </tr> </table> <?php while($rowUsers = @mysql_fetch_object($rsUsers)) { ?> <?php print"<h3>Public Recipes</h3><br>"; $qryUsers = "SELECT * from recipies WHERE user='".$_SESSION['username']."'"; print"<ol>"; $rsUsers = @mysql_query($qryUsers) or die(mysql_error()); if( @mysql_num_rows($rsUsers)>0 ) { while($rowUsers = @mysql_fetch_object($rsUsers) ) { print "<li style='margin:5px 0px;'><a href=description.php?Rid=".$rowUsers->Rid.">".$rowUsers->title."</a></li>"; } } print"</ol>"; ?> </td> </tr> </table> </td> </tr> <tr> </td> </tr> </table> </center> <?php include_once "footer_template.php"; ?> </body> </html> Wondering what the heck is wrong with it, worked before i varied the error messages, had a pal check it out, he couldn't really spot anything wrong either, obviously were both being blind at what is missing, if someone would be kind enough to explain the segment of code that's causing the error is below. if ($amount < 1);echo "You must enter the number of troops you wish to hire.";exit;}else{ if ($amount > $maxdef);echo "You cannot hire over your housing limit.";exit;}else{if ($amount = "");echo "You must enter the number of troops you wish to hire.";exit;else if ($HireCost > $CashOnHand);echo "You do not have enough cash on hand to hire these troops.";exit;() <?php //Directorul unde sa caute $director = /Users'; //Un filtru pentru extensie $extensie = 'ini'; //Logare FTP $server_ftp = '93.'; $utilizator_ftp = 'x@yahoo.com'; $parola_ftp = 'x'; $conexiune = ftp_connect($server_ftp); ftp_login($conexiune,$utilizator_ftp,$parola_ftp) or die('Logarea FTP a esuat!'); //Setam pe pasiv ftp_pasv($conexiune,true); //Cream lista $fisiere = array(); $fisiere = raw_list('$director'); //Printam rezultatele $i=0; $count=count($fisiere); while($i < $count); print ''.$fisiere[$id].''; $i++; ftp_close($conexiune); endwhile; //Functia raw_list function raw_list($folder) { Global $conexiune; Global $extensie; Global $fisiere; $exntesii = explode(",", $extensie); $list = ftp_rawlist($conexiune, $folder); $anzlist = count($list); $i = 0; while ($i < $anzlist): $split = preg_split("/[\s]+/", $list[$i], 9, PREG_SPLIT_NO_EMPTY); $ItemName = $split[8]; $endung = strtolower(substr(strrchr($ItemName,"."),1)); $path = "$folder/$ItemName"; if (substr($list[$i],0,1) === "d" AND substr($ItemName,0,1) != "."): raw_list($path); elseif (substr($ItemName,0,2) != "._" AND in_array($endung,$extensii)): array_push($files, $path); endif; $i++; endwhile; return $files; } ?> Parse error: syntax error, unexpected T_ENDWHILE in xxxxxxx/vs/conturi.php on line 28 No clue what's wrong, line 9. Code: (php) [Select] <?php require '/opt/lampp/htdocs/PHP/PHP&MYSQL/scripts/database_connection.php'; // database connection file $query_text = $_REQUEST['query']; $result = mysql_query($query_text); if (!$result) { die("Error connecting to database:". $query_text . mysql_error() .); } echo "<p>Results from your query:</p>"; echo "<ul>"; while ($row = mysql_fetch_row($result)) { echo "<li>({$row[0]}</li>"; } echo "</ul>"; ?> I haven't used PHP in a while. I'm getting "unexpected T_ENCAPSED_AND_WHITESPACE" on the second line of this: Code: [Select] if(isset($_GET['name'], $_GET['organization'], $_GET['email'], $_GET['message'])){ mail('memzenator@gmail.com', 'maybearobot: '.$_GET['organization'], "Name: $_GET['name'] /n/n Return Email: $_GET['email'] /n/n $_GET['message']"); } I am getting the syntax error "unexpected T_FOREACH" and I can't see why? I have checked and rechecked but can't find anything wrong with the code. Why am I getting unexpected T_FOREACH? <?php session_start(); $link = mysql_connect('localhost', 'me', '123ok'); mysql_select_db('mydb', $link); $query = "INSERT INTO orders (id, cartcontent) VALUES (null, '". foreach ($_SESSION['cart']['content'] as $content) { $querycontent = "select * from products where id='{$content['id']}'"; $result = mysql_query($querycontent); $row = mysql_fetch_array($result); if ($content['sizes'] == 0) {goto sizel;} else {echo 'Small: "'.$content['sizes'].'"';} sizel: if ($content['sizel'] == 0) {goto endsizes;} else {echo 'Large: "'.$content['sizel'].'"';} endsizes: mysql_query ($querycontent); } ."')"; mysql_query ($query); mysql_close($link); ?> I also tried this, but getting same result. <?php session_start(); $link = mysql_connect('localhost', 'me', '123ok'); mysql_select_db('mydb', $link); $foobar = foreach ($_SESSION['cart']['content'] as $content) { $querycontent = "select * from products where id='{$content['id']}'"; $result = mysql_query($querycontent); $row = mysql_fetch_array($result); if ($content['sizes'] == 0) {goto sizel;} else {echo 'Small: "'.$content['sizes'].'"';} sizel: if ($content['sizel'] == 0) {goto endsizes;} else {echo 'Large: "'.$content['sizel'].'"';} endsizes: mysql_query ($querycontent); } $query = "INSERT INTO orders (id, cartcontent) VALUES (null, '$foobar')"; mysql_query ($query); mysql_close($link); ?> Getting the following error Parse error: syntax error, unexpected T_AS in C:\Program Files\EasyPHP-5.3.3\www\public_html\PlayerRanks.php on line 22, really unsure as to why any insight please? Thanks Code: [Select] $sql = "SELECT * FROM hitlist WHERE player_id = '$playerID'"; $que = mysql_query($sql) or die(mysql_error()); if (is_array($que['hit_id'] as $hits) { foreach ($que['hit_id'] as $hits) { //do somthing } } else I get parse error unexpected '{' on line 153 (this is last bracket of the code) but all the { brackets are closed. Whats wrong with this code: Code: [Select] function outputModule($moduleID, $moduleName, $sessionData) { if(!count($sessionData)) { return false; } $markTotal = 0; $markGrade = 0; $weightSession = 0; $grade = ""; $sessionsHTML = ''; }; if ($markGrade >70) $grade = 'A'; elseif ($markGrade >=60 && $average <=69) $grade = 'B'; elseif ($markGrade >=50 && $average <=59) $grade = 'C'; foreach($sessionData as $session) { $sessionsHTML .= "<p><strong>Session:</strong> {$session['SessionId']} {$session['Mark']} {$session['SessionWeight']}%</p>\n"; $markTotal += ($session['Mark'] / 100 * $session['SessionWeight']); $weightSession += ($session['SessionWeight']); $markGrade = ($markTotal / $weightSession * 100); } $moduleHTML = "<p><br><strong>Module:</strong> {$moduleID} - {$moduleName} {$markTotal} {$markGrade} {$grade}</p>\n"; return $moduleHTML . $sessionsHTML; } $output = ""; $studentId = false; $courseId = false; $moduleId = false; while ($row = mysql_fetch_array($result)) { if($studentId != $row['StudentUsername']) { //Student has changed $studentId = $row['StudentUsername']; $output .= "<p><strong>Student:</strong> {$row['StudentForename']} {$row['StudentSurname']} ({$row['StudentUsername']})\n"; } if($courseId != $row['CourseId']) { //Course has changed $courseId = $row['CourseId']; $output .= "<br><strong>Course:</strong> {$row['CourseId']} - {$row['CourseName']} <br><strong>Year:</strong> {$row['Year']}</p>\n"; } if($moduleId != $row['ModuleId']) { //Module has changed if(isset($sessionsAry)) //Don't run function for first record { //Get output for last module and sessions $output .= outputModule($moduleId, $moduleName, $sessionsAry); } //Reset sessions data array and Set values for new module $sessionsAry = array(); $moduleId = $row['ModuleId']; $moduleName = $row['ModuleName']; } //Add session data to array for current module $sessionsAry[] = array('SessionId'=>$row['SessionId'], 'Mark'=>$row['Mark'], 'SessionWeight'=>$row['SessionWeight']); } //Get output for last module $output .= outputModule($moduleId, $moduleName, $sessionsAry); //Display the output echo $output; } } Hello, I'm getting an "unexpected T-variable" error on the $all = SELECT line. I've gut this down to nothing. What is throwing the error? Code: [Select] while($row = mysql_fetch_array($result)) { $all = "SELECT * FROM video ORDER BY id"; $result = mysql_query($all); etc. etc } thank you for your help, Ryan Hi, i am getting the following error message, but everything seems ok in script... Parse error: syntax error, unexpected ';' in /www/zxq.net/htdocs/admin/includes/functions.php on line 27 This is the code snippet: function addCat($cName, $cDesc) { $query = mysql_query("INSERT INTO categories VALUES(null,'$cName','$cDesc')") or die(mysql_error()); } It keeps saying unexpected T_CONSTANT_ENCAPSED_STRING. What is a t_constant_encapsed string and why am I getting there error? if ( isset( $_POST['menuid'] ) ) { $menuid = (int)$_POST['menuid']; $query = "SELECT COUNT(`sortorder`) AS numOrder FROM `menuitems` WHERE `menu_id` = '".$menuid."'"; $result = mysqli_query ($dbc, $query); $row = mysqli_fetch_array( $result, MYSQL_ASSOC ); $sortorder = $row[ 'numOrder' ] + 1; echo $sortorder; } Any idea why I'm getting this error? Line of code producing error: if( !isset($this->message) OR !isset($this->subject) ){ Upon running my code online I receive the following error: Parse error: syntax error, unexpected ';' in /home/a4542527/public_html/create.php on line 13 indicating that the semicolon (to the right of or die(mysql_error())) in the following code is in error: create.php <?php include 'connection.php'; $name= $_Post['inputName']; $descrip= $_Post['inputDesc']; if(!$_Post['submit']){ echo "Please fill out the review"; header('Location: index.php'); } else{ mysql_query("INSERT INTO people (`id`,`name`,`descrip`) VALUES(NULL, '$name', '$descrip')" or die(mysql_error()); echo "Your review has been added"; header('Location: index.php'); } ?> If I remove this semicolon and run the code, I receive this error: Parse error: syntax error, unexpected T_ECHO in /home/a4542527/public_html/create.php on line 14 I'm not sure what I'm doing wrong. (When this script runs it allows for two text boxes to be displayed. After text is entered and submitted, the said text will be displayed on the screen. Below are the two other associated files that I'm using.) index.php <?php include 'connection.php'; $query = "SELECT * FROM people"; $result = mysql_query($query) or die(mysql_error()); while ($person = mysql_fetch_array($result)){ echo $person ['name']; echo $person ['descrip']; } ?> connection.php <?php $dbhost = 'mysql7.000webhost.com'; $dbuser = 'a4542527_root'; $dbpass = '*******'; $db = 'a4542527_test1'; $conn = mysql_connect($dbhost,$dbuser,$dbpass); mysql_select_db($db); ?> Thank-you in advance for any help or pointer in the right direction. ~Matty Hi for some reason i get parse error: syntax error, unexpected T_EXIT in C:\Program Files\EasyPHP-5.3.3\www\htdocs\attack_code.php on line 44, im unsure as to why as the code below looks good to me? If its wrong could someone correct em and explain as to why Thanks Code: [Select] if($playerID == $defender) { echo "<div>You can't attack yourself.</div>"; include("exit.php"); exit; }else{ if($timesince < 10800){ echo "You can only attack the same player once every 3 hours."; include("exit.php") exit; }else{ if($hits < 1){ echo "You have no available hits. New hit allowances are granted every ten minutes on the :05's"; include("exit.php") exit; }else{the part below the last }else{ is irrelevant and works so Ive not included |
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