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hi my husband is customising a website template and wants to add content over a teransparent image but when he adds the tex it goes underneath how can he do this here is where the image is placed `<div id="homepage" class="clear">` `<section class="main_slider"><img src="images/demo/1200x400.gif" alt="">` `</section>` and a preview of image is in http://swinguk.uk/KEVINWORK/ there is currently no css for the div id home or section class if anyone has any ideas be much appreciated ty in advance jan x

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hi all i have recently purchased a script and im attempting to fix it would anyone know how to solve these 2 errors which appear to be coming from this function [09-Jun-2018 12:40:57 UTC] PHP Warning: Division by zero in /home/public_html/includes/functions.php on line 126 [09-Jun-2018 12:41:19 UTC] PHP Fatal error: Maximum execution time of 30 seconds exceeded in /home/public_html/includes/functions.php on line 135 if(!function_exists('Pagination')){ function Pagination($config=array()){ if($_SESSION['lang']=='L1'){ $firstpage = 'First'; $lastpage = 'Last'; } else{ { $firstpage = 'Trang đầu'; $lastpage = 'Trang cuối'; } } $output = ''; ////FIRST ERROR $rs_maxpage = $config['js_numrows_page']>0?ceil($config['js_numrows_page']/$config['per_page']):0; $eitherside = ($config['showeachside'] * $config['per_page']); $paga = …

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hi all im getting the following error PHP Notice: Array to string conversion in /home/public_html/includes/database.php(23) : eval()'d code on line 1 in this file <?php $link = @mysqli_connect($hostname, $username, $password, $database); if (!$link) { die('Could not connect !'); exit(); } else{ mysqli_set_charset($link,'utf8'); } $sqlcf = 'select Variable, Value from '.$table_prefix.'site_settings where IsRead = 0'; $qrycf = mysqli_query($link,$sqlcf); if(!$qrycf) die('Can not load config data !'); elseif(mysqli_num_rows($qrycf)>0){ while($rowcf = mysqli_fetch_assoc($qrycf)){ if('filetype'==trim(strtolower($rowcf['Variable']))) $$rowcf['Variable'] = explode(',', $rowcf['Value']); elseif('numof_record_perpage'==trim(strtolower($rowcf['Variable']))) $$rowcf['Variable'] = intval($rowcf['Value']); elseif('default_language'==trim(strtolower($rowcf['Variable']))) $_SESSION['lang'] = isset($_SESSION['lang'])?$_SESSION['lang']:$rowcf['Value']; else eval("$\$rowcf['Variable'] = \"$rowcf[Value]\";"); } } else die('Can not find setting table !'); and im not sure what i …

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hi all im attempting to update coryApp script and im getting the following error PHP Warning: Division by zero can anyone help on this please be much appreicated heres is where the error is coming from if(isset($error) && !empty($error)) echo '<p style="margin:0px; padding:5px 20px"><font color="#FF0000"><small><i>'.$error.'</i></small></font></p>'; $config['showeachside'] = 4; $config['per_page'] = 20; $config['js_numrows_page'] = mysqli_num_rows(getListState($str, '')); $config['curpage'] = empty($_GET['p'])?1:$_GET['p']; $config['rs_start'] = ($config['curpage']*$config['per_page'])-$config['per_page']; if($config['js_numrows_page'] < $config['per_page']) $config['per_page'] = $config['js_numrows_page']; $page = (isset($_GET['p']) && intval($_GET['p'])>0)?'&p='.$_GET['p']:''; $config['cururl'] = $base_url.'admincp/states.php'.$pstr; $rs_maxpage = ceil($config['js_numrows_page']/$config['per_page']); $paging = Pagination($config); $list = getListState($str, " limit ".$config['rs_start'].", ".$config['per_page']); if(mysqli_num_rows($list)>0){ and here is the function for it if(!function_exists('getListState')){ function getListState($str='', $limit=''){ …

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hi im attempting to update a script from coryApp and im getting the following error PHP Notice: Undefined offset: 11 here is the code where it is coming from can someone point me in the right direction on this ty jan x <td width="15%" class="headrows4" valign="top" align="left"> THIS LINE >>><p><input type="checkbox" style="margin-left:0px; padding-left:0px" value="<?php echo $list['PhotoID'][$i+5];?>" name="imgs[]" /> <?php echo $list['ProfileName'][$i+5]; $extend = ($list['PrimaryPhotoID'][$i+5]==$list['PhotoID'][$i+5])?'Primary':date('m-d-Y', strtotime($list['InsertDate'][$i+5])); echo ' <i>('.$extend.')</i>';?></p> <img src="<?php echo $base_url.$uploaddir.'/'.$list['UserID'][$i+5].'u'.$list['PhotoID'][$i+5].'.'.$list['PhotoExtension'][$i+5];?>" border="0" width='170' /> </td>

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hi ive done a piece of code for if select a region the postcodes pop up for that region all works on a seperate file but if i put on my working script it doesnt work any help would be much appreciated here is test code <!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" xml:lang="en" lang="en"> <head> <meta http-equiv="content-type" content="text/xhtml; charset=utf-8" /> <title>Dynamic Select Statements</title> <script type="text/javascript"> //<![CDATA[ // array of possible countries in the same order as they appear in the event_postalcode selection list var postcodeLists = new Array(4) postcodeLists["empty"] = ["Select a Region"]; postcodeLists["Channel Islands"] = …

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hi im working on a php script and i have come across the following error does anyone have a ideal what can cause this the error is: PHP Warning: count(): Parameter must be an array or an object that implements Countable in /home/letsswin/public_html/dating_V1/dating/admincp/educations.php on line 74. the code where it comes from is : <?php if(isset($error) && !empty($error)) echo '<p style="margin:0px; padding:5px 20px"><font color="#FF0000"><small><i>'.$error.'</i></small></font></p>'; $list = getListed(array('EducationID' => 'Id', 'L1Education' => 'L1Value', 'L2Education' => 'L2Value'), array($table_prefix.'educations', 'Id')); if(count($list)>0){ ?> any help would be much appreciated ty jan x

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hi im attempting to update a php srcipt and fixing errors and i have a error in the following block but cannot work out where if($_SERVER['REQUEST_METHOD']=='POST' && isset($_POST['smcomplete'])){ if(empty($_POST['abouth'])) $err_h = $aboutherorhis; elseif(strlen($_POST['abouth'])<6) $err_h = $datashort; else{ ERROR ON THIS LINE>>> $match = ($_POST['matchsame']=='on')?1:0;slBodyType'], $_POST['slHeightfrom'], $_POST['slHeightt $data = array($_POST['rdgender'], $_POST['slagefrom'], $_POST['slageto'], $_POST['slDatingInterest'], mysqli_real_escape_string($link,$_POST['abouth']), $_POST['slMarital'], $_POST['slReligion'], $_POST['slEducation'], $_POST['o'], $_POST['slSmoking'], $_POST['slDrinking'], $match); if(!yourmatch($data, $_SESSION['user_temp'])) $error = $errordata; else{ $_SESSION['memberid'] = $_SESSION['user_temp']; $_SESSION['memberemail'] = $_SESSION['email_temp']; unset($_SESSION['user_temp']); unset($_SESSION['email_temp']); mysqli_close($link); header('Location: '.$base_url.'members/myaccount.php'); exit(); } } } Ive attempted to put $_POST['slHeightt']; in but it doesnt solve error any ideas or solutions much appreciated many thanks …

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hi all im trying to update a script i found on a disk for a member site and im struggling with the following error PHP Notice: Array to string conversion in /home/public_html/dating_V1/dating/includes/database.php(20) : eval()'d code on line 1 heres the code for that section of php <?php $link = mysqli_connect($hostname, $username, $password, $database); if (!$link) { die('Could not connect !'); exit(); } $sqlcf = 'select Variable, Value from '.$table_prefix.'site_settings where IsRead = 0'; $qrycf = mysqli_query($link,$sqlcf); if(!$qrycf) die('Can not load config data !'); elseif(mysqli_num_rows($qrycf)>0){ while($rowcf = mysqli_fetch_array($qrycf)){ if('filetype'==trim(strtolower($rowcf['Variable']))) $$rowcf['Variable'] = explode(',', $rowcf['Value']); elseif('numof_record_perpage'==trim(strtolower($rowcf['Variable']))) $$rowcf['Variable'] = intval($rowcf['Value']); elseif('default_language'==trim(strtolower($rowcf['Variable']))) $_SESSION['lang'] = isset($_SESSION['lang'])?$_SESSION['lang']:$rowcf['Value']; …

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Hi i have several user types i.e female, male couple etc and when they register each group goes to there own page for profile setup but when they login they go to one standard profile page that came with script how would i redirect them to there own profile page that goes with there group that they sign up into heres code to header them to profile setup page that ive already done which works if($_POST['gender']=='Couple MF') { header('location:registration2MF.php'); } else if($_POST['gender']=='Couple MM') { header('location:registration2MM.php'); } else if($_POST['gender']=='Couple FF') { header('location:registration2FF.php'); } else if($_POST['gender']=='Male') { header('location:registration2M.php'); } else if($_POST['gender']=='Female') { …

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hi all im attempting to update some script and i have a undefined error in a page see coding for error $column = "select * "; $sql .= " from favorites where user_id = '".$_SESSION['userid']."' "; of which the information a want to echo shows up so when i put the following in if(!empty($sql)) { $sql .= " from favorites where user_id = '".$_SESSION['userid']."' "; } the information echoed no longer shows up here is the full code for this <?php include("config/db_connect.php"); include("config/ckh_session.php"); // Add user favorite list if(isset($_POST['favor_id'])) { $check_refer = mysqli_query($conn,"select favor_id , user_id from favorites where favor_id …

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hi all im wondering what i can do to stop this happening advsearch_result.php?city=&country=&gender=a+man&fage=18&lage=99&accomodate=I+can+Accomodate&minheight=152&sexuality=Straight&bodytype=&race=&piercings=No+piercings&tattoos=No+tattoos&dodrink=&smoker=&pincode=&multipleint=&seeking=a+man&meet_type=&travel=I+can+travel&maxheight=213&partner_sexuality=Straight&partner_body_type=Athletic&partner_race=&partner_piercings=No+piercings&partner_tattoos=No+tattoos&partner_drinking=Don't+drink&partner_smoker=&button.x=90&button.y=19 every time i click submit on a form all this come in address bar after file name the code is currently set as $male = $_GET['gender']; $seek = $_GET['seeking']; $age1 = $_GET['fage']; $age2 = $_GET['lage']; etc ... many thanks jan x

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ive just done a simple register form where i can check username exists and before i put the username exists in the register works but when i have the username check in it doesnt work can someone check this over for me the script and error log are as follows: with username check included <?php require 'Connections/connections.php'; ?> <?php if(isset($_POST['register'])) { $username = mysqli_real_escape_string($conn, $_POST['username']; $email = mysqli_real_escape_string($conn, $_POST['email']; $password = mysqli_real_escape_string($conn, $_POST['password']; $storePassword = password_hash($password, PASSWORD_BCRYPT, array('cost' => 10)); //username check start $username = mysqli_query($conn,"SELECT username FROM users WHERE username='$username'"); $count = mysqli_num_rows($username); if($count!=0){ die("Username already exists! Please type …

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Hi im ageeting the following error in my script can anyone find a fix or solution hus jan x Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /home/letsswin/public_html/template_pageTop.php on line 7 Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in /home/letsswin/public_html/template_pageTop.php on line 8 //check for new pm's $pm_n = '<img src="images/pmStill.gif" width="17" height="12" alt="Pm" title="This pm is for logged in members">'; $sql = "SELECT id FROM pm WHERE (receiver='$log_username' AND parent='x' AND rdelete='0' AND rread='0') OR (sender='$log_username' AND sdelete='0' AND parent='x' AND hasreplied='1' AND sread='0') LIMIT 1"; $query = mysqli_query($db_conx, $sql); $numrows = mysqli_num_rows($query); …

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hi all im trying to place this piece of coding in a table that i iam echoing if anyone has any ideas as been trying for last god knows how long $fetch_img1 = mysqli_fetch_array(mysqli_query($conn,"select user_image from user_images where user_id = '".$show_id."' and main_image = '1' ")); $show_image = $fetch_img1['user_image']; ?> <div class="box"> <?php if($show_image != '') { ?> <a href="viewprofile.php?profid=<?php echo $show_id;?>&gen=<?php echo $show_gender;?>"><img src="images/user_images/smallthumb/<?php echo $show_image;?>" border="0" width="100" height="80"/></a> <?php } else { ?> <a href="viewprofile.php?profid=<?php echo $show_id;?>&gen=<?php echo $show_gender;?>"><img src="images/blank.jpg" border="0" width="100" height="80" /></a> <?php } ?> </div> And here is the table im attempting to place it …

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hi ive got a html table echoing through php and im having problems displaying image in the table heres what i have so far if($fetch_image['user_image']!='') { echo '<td>'.$fetch_image['user_image'].'</td>'; } else { echo '<td>'.<img src='images/blank_big.jpg' height='150px;' width='150px;' />.'</td>'; } and here is the fetch section $user_image = mysqli_query($conn,"select * from user_images where user_id = '".$fetch_user['user_id']."' and main_image = '1' "); $fetch_image = mysqli_fetch_array($user_image); and here is the error i am getting PHP Parse error: syntax error, unexpected '<' in any help on this would be much appreciated ty jan x

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good got job nearly finished but one slight glitch i have 2 entries in database but only one is showing twice in table results how can i get to show both results heres code fetch and connection block <?php include("config/db_connect.php"); //include("config/ckh_session.php"); $sql = "SELECT * FROM meets"; $result = mysqli_query($conn, $sql); //fetch user $meets = mysqli_query($conn,"select * from meets"); $fetch_meets = mysqli_fetch_array($meets); $query = mysqli_query($conn,"select * from user"); $fetch_user = mysqli_fetch_array($query); $user_birthdate = $fetch_user['user_birthdate']; $partneruser_birthdate = $fetch_user['partneruser_birthdate']; //fetch user image $user_image = mysqli_query($conn,"select * from user_images where user_id = '".$fetch_user['user_id']."' and main_image = '1' "); $fetch_image = mysqli_fetch_array($user_image); // for …

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hi i have a form for members to fill in and i want there own personal submissions under form where i will be adding options but it will only work if i put the submitted data above form else i get white screen heres my coding **top php block ** <?php include("config/db_connect.php"); //include("config/ckh_session.php"); if (isset($_POST['submit'])) { $event_type = mysqli_real_escape_string($conn, $_POST['event_type']); $event_date = mysqli_real_escape_string($conn, $_POST['event_date']); $event_country = mysqli_real_escape_string($conn, $_POST['event_country']); $event_postcode = mysqli_real_escape_string($conn, $_POST['event_postcode']); $event_title = mysqli_real_escape_string($conn, $_POST['event_title']); $event_description = mysqli_real_escape_string($conn, $_POST['event_description']); $event_ltm = mysqli_real_escape_string($conn, $_POST['event_ltm']); $sql = "INSERT INTO meets (event_type, event_date, event_country, event_postcode, event_title, event_description, event_ltm) VALUES ('$event_type', '$event_date', '$event_country', …

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hi all ive got a file to display info from database and im trying some way to fetch username from another table for it i have put this at top of file. <?php $query = mysqli_query($conn,"select * from user"); $fetch_user = mysqli_fetch_array($query); ?> and here is the section for displaying data <?php $sql = "SELECT * FROM meets"; $result = mysqli_query($conn, $sql); if (mysqli_num_rows($result) > 0) { echo "<table><tr><th>ID</th><th>Event Type</th><th>Event Date</th><th>Event Region</th><th>Event Postcode</th><th>Event Title</th><th>Event Description</th><th>Event Seeking</th></tr>"; // output data of each row while($row = mysqli_fetch_assoc($result)) { echo "<tr><td>".$row["id"]."</td><td>".$row["event_type"]."</td><td>".$row["event_date"]."</td><td>".$row["event_country"]."</td><td>".$row["event_postcode"]."</td><td>".$row["event_title"]."</td><td>".$row["event_description"]."</td><td>".$row["event_ltm"]. "</td></tr>"; } echo "</table>"; } else { echo "0 results"; } mysqli_close($conn); …

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hi im attempting to add mysqli_real_escape_string into form to make more secure and after i added it in code below $event_type = $_POST['event_type']; $event_date = $_POST['event_date']; $event_country = $_POST['event_country']; $event_postcode = mysqli_real_escape_string($conn, $_POST['event_postcode']); $event_title = mysqli_real_escape_string($conn, $_POST['event_title']); $event_description = mysqli_real_escape_string($conn, $_POST['event_description']); $event_ltm = $_POST['event_ltm']; like so and go to fill in form it isnt recording the information to the database but if i remove it the information goes into database is there anything else i have to do like in the form itself to make this work. <br class="clear" /> <label for="event_postcode">postcode</label><input type="text" name="event_postcode" id="event_postcode" value="<?php if(!empty($event_postcode)) {?><?php echo $_POST['event_postcode']?><?php …

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hi im struggling with this error see section below if anyone can help be much appreciated ty jan x <?php include("config/db_connect.php"); include("config/ckh_session.php"); // Inserting user Details code if(isset($_POST['submitmeet'])) { $event_title = addslashes($_POST['event_title']); $description = addslashes($_POST['description']); $insert = mysqli_query($conn,"update meets set event_type = '".$_POST['event_type']."' , event_date = '".$_POST['event_date']."' , stateid = '".$_POST['stateid']."' , area = '".$_POST['area']."' , event_title = '".$event_title."' , description = '".$description."' , ltm = '".$_POST['ltm']."' where user_id = '".$_SESSION['last_id']."' ") or die(mysqli_error($conn)); if($insert) { header("location: searchmeets.php"); } } // Fetch user details $query = mysqli_query($conn,"select * from user where user_id = '".$_SESSION['last_id']."' "); //THIS LINE IS THE ERROR …

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hi i have the following form where a member chooses how long they want to be be displayed on another page like example below <select name="hours"> <option value="1"<?php if(!empty($hours)) {?><?php if($_POST['hours']=="1") { echo "selected"; } ?><?php }?>>1</option> <option value="3"<?php if(!empty($hours)) {?><?php if($_POST['hours']=="3") { echo "selected"; } ?><?php }?>>3</option> <option value="6"<?php if(!empty($hours)) {?><?php if($_POST['hours']=="6") { echo "selected"; } ?><?php }?>>6</option> <option value="9"<?php if(!empty($hours)) {?><?php if($_POST['hours']=="9") { echo "selected"; } ?><?php }?>>9</option> <option value="12"<?php if(!empty($hours)) {?><?php if($_POST['hours']=="12") { echo "selected"; } ?><?php }?>>12</option> </select> What would i need to put in to make sure a username disappears of the list would it …

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hi im attempting to update a script and i have come across the following error PHP Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in so here was the orginal $sql = "select * from chat where (chat.to = '".mysqli_real_escape_string($_SESSION['usernamechat'])."' AND recd = 0) order by id ASC"; of which i changed to $sql = "select * from chat where (chat.to = '".mysqli_real_escape_string($conn, $_SESSION['usernamechat'])."' AND recd = 0) order by id ASC"; and still getting the same error any ideas anyone ty in advance jan x

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hi ive picked up a script and although ive put database details in its not picking them up heres the database file ommiting my database details of course config.php <?php ob_start(); session_start(); ini_set("display_errors", 0); $hostname = "localhost"; $database = "matureco_test"; $password = "xxxxx"; $username = "xxxx"; and heres the database file <?php $conn = mysqli_connect($hostname, $username, $password,$database); if (!$conn) { die('Could not connect !'); exit(); } else{ mysqli_set_charset('utf8',$conn); } $sqlcf = 'select Variable, Value from '.$table_prefix.'site_settings where IsRead = 0'; $qrycf = mysqli_query($conn,$sqlcf); if(!$qrycf) die('Can not load config data !'); elseif(mysqli_num_rows($qrycf)>0){ while($rowcf = mysqli_fetch_array($qrycf)){ if('filetype'==trim(strtolower($rowcf['Variable']))) $$rowcf['Variable'] = explode(',', $rowcf['Value']); elseif('numof_record_perpage'==trim(strtolower($rowcf['Variable']))) …

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hi i have the following error [09-Nov-2016 16:22:28 America/Chicago] PHP Notice: Undefined index: HTTP_REFERER in /home/matureco/public_html/index.php on line 224 [09-Nov-2016 16:22:28 America/Chicago] PHP Notice: Undefined index: email in /home/matureco/public_html/index.php on line 225 ive checked and they are in the php block see code below can anyone tell me why this is happening regards jan x <td><input type="hidden" name="go" id="go" value="<?php echo $_SERVER['HTTP_REFERER'];?>" /> <input type="text" name="email" id="email" size="30" value="<?php if(!empty($email)) {?><?php echo $_POST['email']?><?php }?>" maxlength="20"></td> $email = $_POST['email']; $password = md5($_POST['pass']); $go = $_POST['go']; $_SESSION['SESS_email'] = $logmysqli_data['user_email']; $lastlog = mysqli_query( $conn,"update user_lastlogin set lastlogin = '$sdate' , ipaddress = '".$_SERVER['REMOTE_ADDR']."' …

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Hi all im currently updating a script and repairing it after a page full or errors most ive done through searching but im stuck on this on if anyone can help be much appreciated jan PHP Notice: Only variables should be passed by reference on line 16 the piece of code for this is if(isset($_POST['sub1'])) { //$fn = date('U').$_FILES['userimage']['name']; $ext = end(explode('.',$_FILES['userimage']['name'])); //THIS IS THE ERROR LINE// $fn = date('U').".".$ext; $ft = $_FILES['userimage']['type']; $fs = $_FILES['userimage']['size']; $ftmp = $_FILES['userimage']['tmp_name']; move_uploaded_file($ftmp, "images/user_images/$fn"); $thumb = new Thumbnail('images/user_images/'.$fn); $thumb->resize(100,100); $thumb->save('images/user_images/smallthumb/'.$fn); $thumb1 = new Thumbnail('images/user_images/'.$fn); $thumb1->resize(405,540); $thumb1->save('images/user_images/bigthumb/'.$fn); mysqli_query($conn,"update user set user_image = '".$fn."' where …

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im attempting to add date of birth and age to a form and database the date of birth works fine and shows in database as it should be but the age is showing 0 in age column ive tried different things after searching and its staying the same heres the code for age and date of birth in the php block $year = $_POST['year']; $month = $_POST['month']; $day = $_POST['day']; $DOB = $year . "-" . $month . "-" . $day; $age = $_POST['age']; And here is the database string mysqli_query($conn,"INSERT INTO signup (f_name,email,password,DOB,age,memtype) VALUES ('$fname','$email','$EncryptPassword','$DOB','$age','$memtype')"); date of birth is …

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hi all im trying to update a script and i have the following errors which ive googled and im lost if anyone can help be much appreciated ty jan x PHP Notice: Use of undefined constant prepare_insert - assumed 'prepare_insert' in /home/matureco/public_html/UPLOADIMAGE/config/db_connect.php on line 32 code for this error is below //prepare to insert data if(! function_exists(prepare_insert) ) { function prepare_insert($data) { $data = trim ( $data ); if (get_magic_quotes_gpc ()) $data = stripslashes ( $data ); return mysqli_escape_string ( $data ); } } PHP Notice: Use of undefined constant prepare_show - assumed 'prepare_show' in /home/matureco/public_html/UPLOADIMAGE/config/db_connect.php on line 43 if(! …

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hi all im in the middle of following a tutorial online for a rating system and im encountering a problem when i click a number of 1/5 to rate a topic the rating is not going into the database can someone check the code to see what going off sql tables CREATE TABLE IF NOT EXISTS `articles` ( `id` int(11) NOT NULL, `title` varchar(255) COLLATE utf8_unicode_ci NOT NULL ) ENGINE=MyISAM AUTO_INCREMENT=4 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; -- -- Dumping data for table `articles` -- INSERT INTO `articles` (`id`, `title`) VALUES (1, 'test article one'), (2, 'test article two'), (3, 'test article three'); …

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hi im not sure if come into the right section but im looking at adding a chatroom to my site for my users to use and im wanting it where they dont have to log in or register to chatroom instead they login on site and the username is same as username in site i know theres a lot that you can intergrate into most cms but my site isnt based on cms what chat script would you recommend for this ty in advance kevin

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The End.