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PHP - Question About: Unexpected ';'
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 Similar Tutorialshello 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> Hi, 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> 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.'; } } } ?> 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 $end in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/showroom/inserts.php on line 23 Don't know whats wrong? Am I blind? I don't see the problem with this code. It says: Parse error: syntax error, unexpected '}' in /data/21/2/40/160/2040975/user/2235577/htdocs/edit_user1.php on line 35 Code: [Select] <html> <body> <?php if ( (isset($_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> Hi, could someone please tell me why I'm getting unexpected $end error in the following code? <?php function documentType(){ echo <<<HEREDOC <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" ` ` "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> } HEREDOC; ?> I'm getting an unexpected t_string and I'm not sure where on this line I'm missing it. Code: [Select] <?php echo "<img src=\".base_url()."assets/images/avatar.jpg\".$data->avatar_file."" alt=\"\" />"; ?>
if($a==1)
Can you please help to identify and correct the above code error Syntax error found: unexpected T_ELSEIF on line 23 Code: [Select] <?php require("config.php"); $result = mysql_query("SELECT name, mics, status, date_format(eventStart, '%M %e %Y') as start, date_format(eventFinish, '%M %e %Y') as end FROM events where status='S' or status='A' order BY eventStart"); $sta=$row['status']; { echo "<table rules='rows'> <tr> <th width='50'><font color='yellow' size='1'>MICS</th> <th width='200'><font color='yellow' size='1'>Charity</th> <th width='125'><font color='yellow' size='1'>Start Date</th> <th width='125'><font color='yellow' size='1'>End Date</th> </tr>"; while ($row = mysql_fetch_array($result)) { if ($sta == "A"); { echo "<tr>"; echo "<td><font color='red' size='1'>" . $row['mics'] . "</td>"; echo "<td><font color='red' size='1'>" . $row['name'] . "</td>"; echo "<td><font color='red' size='1'>" . $row['start'] . "</td>"; echo "<td><font color='red' size='1'>" . $row['end'] . "</td>"; echo "</tr>"; } elseif ($sta == "S"); { echo "<tr>"; echo "<td><font color='#ffffff' size='1'>" . $row['mics'] . "</td>"; echo "<td><font color='#ffffff' size='1'>" . $row['name'] . "</td>"; echo "<td><font color='#ffffff' size='1'>" . $row['start'] . "</td>"; echo "<td><font color='#ffffff' size='1'>" . $row['end'] . "</td>"; echo "</tr>"; } } echo "</table>"; } mysql_close($dbh); ?> [code] I don't see it. Thanks What is a T_STRING? google.com cannot find anything about it in the PHP manual. ... and, why am I getting "unexpected T_STRING" for this line? Code: [Select] if ($content['sizes'] == 0) {goto sizem;} else {echo "<B>Small:</B> '.$content['sizes'].'";} 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 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 $output; } } Hi for some reason i get parse error: syntax error, unexpected T_EXIT in C:\Program Files\EasyPHP-5.3.3\www\htdocs\attack_code.php on line 44, im unsure as to why as the code below looks good to me? If its wrong could someone correct em and explain as to why Thanks Code: [Select] if($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($hits < 1){ echo "You have no available hits. New hit allowances are granted every ten minutes on the :05's"; include("exit.php") exit; }else{the part below the last }else{ is irrelevant and works so Ive not included I haven't used PHP in a while. I'm getting "unexpected T_ENCAPSED_AND_WHITESPACE" on the second line of this: Code: [Select] if(isset($_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 $_GET['message']"); } I'm getting a unexpected t_if error and not sure where my fix is. Code: [Select] <?php if($templateVar == '') { $templateVar = 'peach'; //defaulting to be safe } else { $templateVar = $templateVar; //passed from controller } ?> <?php $this->load->view($templateVar . '/header'); ?> <?php $this->load->view($templateVar . '/navigation'); ?> <!-- End Navigation --> <?php $this->load->view($templateVar . '/msgbox'); ?> <?php if((empty($bodyType))||(!isset($bodyType))||(trim($bodyType)=="")){$this->load->view($templateVar . '/body_full');} elseif($bodyType == "full"){$this->load->view($templateVar . '/body_full');} /* Commented out for now, as currently we only have one layout dimension, this is $ out various sections that have different layouts, 2 columns, 3 columns, columns$ on and so forth. elseif($bodyType == "2column"){$this->load->view($templateVar . '/body_2column$ elseif($bodyType == "3column"){$this->load->view($templateVar . '/body_3column$ elseif($bodyType == "blog"){$this->load->view($templateVar . '/body_blog');} */ else{$this->load->view($templateVar . '/body_full');} ?> <?php $this->load->view($templateVar . '/footer'); ?> Issue line: Code: [Select] if((empty($bodyType))||(!isset($bodyType))||(trim($bodyType)=="")){$this->load->view($templateVar . '/body_full');} Hi All Can anyone give me any pointers as to why I'm getting the following error from the code below? Code: [Select] <?php require_once "../config.php"; $sql = 'SELECT value FROM `mdl_scorm_scoes_track` WHERE scormid =338 AND element = "cmi.core.lesson_status"'; $sql = mysql_query($sql); while ($row = mysql_fetch_array($sql)){ switch($row['value']{ case "passed": $passed ++; break; case "incomplete": $incomplete++; break; case "completed": $completed++; break; default: $other++; break; } } echo "Passed = $passed <br/>"; echo "Incomplete = $incomplete <br/>"; echo "Completed = $completed <br/>"; echo "Others = $other<br/>"; ?> Parse error: syntax error, unexpected T_CASE in D:\Documents\AI24\Web\skooch\dccout\index.php on line 8 Cheers in advance guys I am getting this error message:
Parse error: syntax error, unexpected end of file in C:\xampp\htdocs\sitepoint\preupload.php on line 67with this script:
<?php
include 'config.inc.php';
// initialization
$photo_upload_fields = '';
$counter = 1;
// If we want more fields, then use, preupload.php?number_of_fields=20
$number_of_fields = (isset($_GET['number_of_fields'])) ?
(int)($_GET['number_of_fields']) : 5;
// Firstly Lets build the Category List
$result = mysql_query('SELECT category_id,category_name FROM gallery_category');
while($row = mysql_fetch_array($result)) {
$photo_category_list .= <<<__HTML_END
<option value="$row[0]">$row[1]</option>n
__HTML_END;
}
mysql_free_result( $result );
// Lets build the Image Uploading fields
while($counter <= $number_of_fields) {
$photo_upload_fields .= <<<__HTML_END
<tr><td>
Photo {$counter}:
<input name="photo_filename[]"
type="file" />
</td></tr>
<tr><td>
Caption:
<textarea name="photo_caption[]" cols="30"
rows="1"></textarea>
</td></tr>
__HTML_END;
$counter++;
}
// Final Output
echo <<<__HTML_END
<html>
<head>
<title>Lets upload Photos</title>
</head>
<body>
<form enctype="multipart/form-data"
action="upload.php" method="post"
name="upload_form">
<table width="90%" border="0"
align="center" style="width: 90%;">
<tr><td>
Select Category
<select name="category">
$photo_category_list
</select>
</td></tr>
<! - Insert the image fields here -->
$photo_upload_fields
<tr><td>
<input type="submit" name="submit"
value="Add Photos" />
</td></tr>
</table>
</form>
</body>
</html>
__HTML_END;
?>
on line 67It points to the closing tag: ?>Does anybody have a suggestion how to solve this issue? 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; } |
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